# Light Waves

## Concept Of Physics

### H C Verma

1   Find the range of frequency of light that is visible to an average human being $(400 nm < \lambda < 700 nn)$.

##### Solution :

$\frac{1}{7 \times 10^{-1}} < \frac{1}{\lambda} < \frac{1}{4 \times 10^{-7}} \Rightarrow \frac{3 \times 10^{8}}{7 \times 10^{-7}} < \frac{c}{\lambda} < \frac{3 \times 10^{8}}{4 \times 10^{-7}}$ (Where, $c$ = speed of light = $3 \times 10^{8}\ m/s$)

Given that, $400 m < \lambda < 700 nm$ $\\$ $\frac{1}{700\ nm} < \frac{1}{\lambda} < \frac{1}{400\ nm}$

$\Rightarrow 4.3 \times 10^{14} < \frac{c}{\lambda} < 7.5 \times 10^{14}$ $\\$ $\Rightarrow 4.3 \times 10^{14}\ Hz < f < 7.5 \times 10^{14}\ Hz$.

2   The wavelength of sodium light in air is $589\ nm$. (a) Find its frequency in air. (b) Find its wavelength in water $(refractive\ index = 1.33)$. (c) Find its frequency in water. (d) Find its speed in water.

##### Solution :

$\frac{\mu_{a}}{\mu_{w}} = \frac{V_{w}}{V_{a}} \Rightarrow = v_{w} = \frac{\mu_{a}V_{a}}{\mu_{w}} = \frac{3 \times 10^{8}}{1.33} = 2.25 \times 10^{8}\ m/s$

a) $f_{a} = \frac{3 \times 10^{8}}{589\times 10^{-9}} = 5.09 \times 10^{14}\ sec^{-1} \big[\because f = \frac{c}{\lambda}\big]$

$f_{w} = f_{a} = 5.09 \times 10^{14} sec^{–1}$ [Frequency does not change]

Given that, for sodium light, $\lambda = 589\ nm = 589\times 10^{–9} m$

$\frac{\mu_{a}}{\mu_{w}} = \frac{\lambda_{w}}{\lambda_{a}} \Rightarrow \frac{1}{1.33} = \frac{\lambda_{w}}{589 \times 10^{-9}} \Rightarrow \lambda_{w} = 443\ nm$

3   The index of refraction of fused quartz is $1\cdot472$ for light of wavelength $400\ nm$ and is $1\cdot452$ for light of wavelength $760\ nm$. Find the speeds of light of these wavelengths in fused quartz.

##### Solution :

So, $\frac{1472}{1} = \frac{3 \times 10}{v_{400}} \Rightarrow v_{400} = 2.04 \times 10^{8}\ m/sec$

Again, $\frac{1472}{1} = \frac{3 \times 10}{v_{760}} \Rightarrow v_{760} = 2.07 \times 10^{8}\ m/sec$

We know that, $\frac{\mu_{2}}{\mu_{1}} = \frac{V_{1}}{V _{2}}$

[because, for air, $\mu = 1$ and $v = 3\times 10^{8}\ m/sec$].

4   The speed of the yellow light in a certain liquid is $2.4 \times 10^{8}\ m/s$. Find the refractive index of the liquid.

##### Solution :

$\mu_{t} = \frac{1 \times 3\times 10^{8}}{(2.4)\times 10^{8}} = 1.25\ \big[\ since, \mu = \frac{velocity\ of\ light\ in\ vaccum}{velocity\ of\ light\ in\ the\ given\ medium}\big]$

5   Two narrow slits emitting light in phase are separated by a distance of $1\cdot0\ cm$. The wavelength of the light is $5\cdot0 \times 10^{-7}\ m$. The interference pattern is observed on a screen placed at a distance of $1\cdot0\ m$. (a) Find the separation between the consecutive maxima. Can you expect to distinguish between these maxima ? (b) Findthe separation between the sources which will give a separation of $1\cdot0\ mm$ between the consecutive maxima.

##### Solution :

$10^{–3}\ m = \frac{5 \times 10^{-7} \times 1}{D} \Rightarrow D = 5 \times 10^{-4}\ m = 0.50\ mm$.

a) Separation between two consecutive maxima is equal to fringe width.

b) When, $\beta = 1\ mm = 10^{–3}\ m$

Given that, $d = 1\ cm = 10^{–2}\ m,\ \lambda = 5 \times 10^{–7}\ m$ and $D = 1\ m$

So, $\beta = \frac{\lambda{D}}{d} = \frac{5 \times 10^{-7} \times 1}{10^{-2}}\ m = 5 \times 10^{-5}\ m =0.05\ mm$.

6   The separation between the consecutive dark fringes in a Young's double slit experiment is $1\cdot0\ mm$. The screen is placed at a distance of $2\cdot5\ m$ from the slits and the separation between the slits is $1\cdot0\ mm$. Calculate the wavelength of light used for the experiment.

##### Solution :

So,$10^{-3}\ m = \frac{25 \times \lambda}{10^{-3}} \Rightarrow \lambda = 4 \times 10^{-7}\ m = 400\ nm$.

Given that, $\beta = 1\ mm = 10^{-3}\ m, D = 2.t\ m$ and $d = 1\ mm = 10^{-3}\ m$.

7   In a double slit interference experiment, the separation between the slits is $1\cdot0\ mm$, the wavelength of light used is $5\cdot0\ \times 10^{-7}\ m$ and the distance of the screen from the slits is $1\cdot0\ m$. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimeter width on the screen ?

##### Solution :

Given that, $d = 1\ mm = 10^{–3}\ m, D = 1\ m$ $\\$ So, fringe with = $\frac{D\lambda}{d} = 0.5\ mm$. $\\$ a) So, distance of centre of first minimum from centre of central maximum = $\frac{0.5}{2}\ mm = 0.25\ mm$ $\\$ b) No. of fringes = $\frac{10}{0.5} = 20$.

8   In a Young's double slit experiment, two narrow vertical slits placed $0\cdot800\ mm$ apart are illuminated by the same source of yellow light of wavelength $589\ nm$. How far are the adjacent bright bands in the interference pattern observed on a screen $2\cdot00\ m$ away ?

##### Solution :

Given that, $d = 0.8\ mm = 0.8\times 10^{–3}\ m, \lambda = 589\ nm = 589 \times 10^{–9} m$ and $D = 2\ m$. $\\$ So, $\beta = \frac{D\lambda}{d} = \frac{589 \times 10^{-9}\times 2}{0.8 \times 10^{-3}} = 1.47 \times 10^{-3}\ m = 147\ mm.$

9   Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength $500\ nm$. The separation between the slits is $2\cdot0\ \times 10^{-3}\ M$.

##### Solution :

As shown in the figure, angular separation $\theta = \frac{\lambda{D}}{dD} = \frac{\lambda}{d}$

$= 25 \times 10^{-5}\ radian = 0.014\ degree$.

Given that, $\lambda = 500\ nm = 500 \times 10^{–9}\ m$ and $d = 2 \times 10^{–3}\ m$

So, $\theta = \frac{\beta}{D} = \frac{\lambda}{d} = \frac{500 \times 10^{-9}}{2 \times 10^{-3}} = 250 \times 10^{-6}$

10   A source emitting light of wavelengths $480\ nm$ and $600\ nm$ is used in. a double slit interference experiment. The separation between the slits is $0\cdot25\ mm$ and the interference is observed on a screen placed at $150\ cm$ from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.

##### Solution :

$7^{th}$ bright fringe of violet light overlaps with $4^{th}$ bright fringe of red light (minimum). Also, it can be seen that $14^{th}$ violet fringe will overlap 8^{th} red fringe. $\\$ Because, $m/n = 7/4 = 14/8$.

So, $y_1 = \frac{D\lambda_1}{d} = \frac{1.5 \times 480 \times 10^{-9}}{0.25 \times 10^{-3}} = 2.88\ mm$ $\\$ $y_2 = \frac{1.5 \times 600 \times 10^{-9}}{0.25 \times 10^{-3}} = 3.6\ mm$

Let $m^{th}$ bright fringe of violet light overlaps with $n^{th}$ bright fringe of red light. $\\$ $\therefore$ $\frac{m \times 400\ nm \times {D}}{d} = \frac{n \times 700\ nm \times D}{d} \Rightarrow \frac{m}{n} = \frac{7}{4}$

We know that, the first maximum (next to central maximum) occurs at $y = \frac{\lambda{D}}{d}$ $\\$ Given that, $\lambda_1 = 480\ nm, \lambda_2 = 600 nm, D = 150\ cm = 1.5\ m$ and $d = 0.25\ mm = 0.25 \times 10^{-3}\ m$

So, the separation between these two bright fringes is given by, $\\$ $\therefore$ separation $= y_2\ –\ y_1 = 3.60\ –\ 2.88 = 0.72\ mm$.

11   White light is used in a Young's double slit experiment. Find the minimum order of the violet fringe $(\lambda\ =\ 400\ nm)$ which overlaps with a red fringe $(\lambda\ =\ 700\ nm)$.

##### Solution :

$7^{th}$ bright fringe of violet light overlaps with $4^{th}$ bright fringe of red light (minimum). Also, it can be seen that $14^{th}$ violet fringe will overlap $8^{th}$ red fringe. $\\$ Because, $m/n = 7/4 = 14/8$.

Let $m^{th}$ bright fringe of violet light overlaps with $n^{th}$ bright fringe of red light. $\\$ $\therefore$ $\frac{m \times 400\ nm \times {D}}{d} = \frac{n \times 700\ nm \times D}{d} \Rightarrow \frac{m}{n} = \frac{7}{4}$

12   Find the thickness of a plate which will produce a change in optical path equal to half the wavelength $\lambda$ of the light passing through it normally. The refractive index of the plate is $\mu$.

##### Solution :

$\Rightarrow t = \frac{\lambda}{2(\mu - 1)}$

Let, $t$ = thickness of the plate $\\$ Given, optical path difference $= (\mu - 1)t = \frac{\lambda}{2}$

13   A plate of thickness $t$ made of a material of refractive index $\mu$ is placed in front of one of the slits in a double slit experiment. (a) Find the change in the optical path due to introduction of the plate. (b) What should be the minimum thickness t which will make the intensity at the centre of the fringe pattern zero ? Wavelength of the light used is $\lambda$. Neglect any absorption of light in the plate.

##### Solution :

b) To have a dark fringe at the centre the pattern should shift by one half of a fringe.

a) Change in the optical path $= \mu{t} - t = (\mu - 1)t$

$\Rightarrow (\mu - 1)t = \frac{\lambda}{2} \Rightarrow t = \frac{\lambda}{2(\mu - 1)}$

14   A transparent paper (refractive index = $1\cdot45$) of thickness $0\cdot02\ mm$ is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength $620\ nm$. How many fringes will cross through the centre if the paper is removed ?

##### Solution :

We know, when the transparent paper is pasted in one of the slits, the optical path changes by $(\mu – 1)t$. $\\$ Again, for shift of one fringe, the optical path should be changed by $\lambda$. $\\$ So, no. of fringes crossing through the centre is given by,

Given that, $\mu = 1.45,\ t\ = 0.02\ mm = 0.02 \times 10^{-3}\ m$ and $\lambda = 620\ nm = 620 \times 10^{-9}\ m$

$n = \frac{(\mu - t)}{\lambda} = \frac{0.45 \times 0.02 \times 10^{-3}}{620 \times 10^{-9}} = 14.5$

15   In a Young's double slit experiment using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index $1\cdot6$ and thickness $1\cdot964\ micron\ (1\ micron = 10^{-6}\ m)$ is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.

##### Solution :

Again, when the distance between the screen and the slits is doubled, $\\$ Fringe width $= \frac{\lambda(2D)}{d} \ \ \ \ \ \ \ \ \ \ \ .....(2)$

We know, number of fringes shifted $= \frac{(\mu - 1)}{\lambda}$

$\Rightarrow = \frac{(\mu - 1)t}{\lambda} = \frac{(1.6 - 1)\times (1.964)\times 10^{-6}}{2} = 589.2\times 10^{-9} = 589.2\ nm.$

$= \frac{(\mu - 1)t}{\lambda} \times \frac{\lambda{D}}{d} = \frac{(\mu - 1)tD}{d} \ \ \ \ \ \ \ ......(1)$

In the given Young’s double slit experiment, $\\$ $\mu = 1.6 ,\ t = 1.964\ micron = 1.964 \times 10^{-6}\ m$

From (1) and (2), $\frac{(\mu - 1)tD}{d} = \frac{\lambda(2D)}{d}$

So, the corresponding shift = No.of fringes shifted $\times$ fringe width

16   A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is $0\cdot50\ mm$ and the separation between the slits is $0\cdot12\ cm$. The refractive index of mica and polysterene are $1\cdot58$ and $1\cdot55$ respectively for the light of wavelength $590\ nm$ which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width ? (b) At what distance from the centre will the first maximum be located ?

##### Solution :

$\Rightarrow$ There are $25$ fringes and $0.43$ th of a fringe $\\$ $\Rightarrow$ There are $13$ bright fringes and $12$ dark fringes and $0.43$ th of a dark fringe. $\\$ So, position of first maximum on both sides will be given by $\\$ $\therefore x = 0.43 \times 4.91 \times 10^{-4} = 0.021\ cm$ $\\$ $x^{'} = (1 - 0.43) \times 4.91 \times 10^{-4} = 0.028\ cm$ (since, fringe width $= 4.91 \times 10^{-4}\ m$)

a) Fringe width $= \dfrac{D\lambda}{d} = \dfrac{1 \times 590 \times 10^{-9}}{12 \times 10^{-4}} = 4.91 \times 10^{-4}\ m.$

So, No. of fringes shifted $= \dfrac{0.015 \times 10 ^{-3}}{590 \times 10^{-3}} = 25.43.$

Given that, $t_1 = t_2 = 0.5 mm = 0.5 \times 10^{-3}\ m \mu_m = 1.58$ and $\mu_p = 1.55$, $\\$ $\lambda = 590\ nm = 590 \times 10^{-9}\ m, d = 0.12\ cm = 12 \times 10^{-4} m, D = 1\ m$

b) When both the strips are fitted, the optical path changes by $\\$ $\Delta{X} = (\mu - 1)t_1 - (\mu_m - \mu_p)t$ $\\$ $= (1.58 - 1.55) \times(0.5)(10^{-3}) = 0.015 \times 10^{-13}\ m.$

17   Two transparent slabs having equal thickness but different refractive indices $\mu_{1}$ and $\mu_{2}$ are pasted side by side to form a composite slab. This slab is placed just after the double slit in a Young's experiment so that the light from one slit goes through one material and the light from the other slit goes through the other material. What should be the minimum thickness of the slab so that there is a minimum at the point $P_{o}$ which is equidistant from the slits ?

##### Solution :

So, $\Rightarrow \dfrac{\lambda}{2} = (\mu_1 - \mu_2)t \Rightarrow t = \dfrac{\lambda}{2(\mu_1 - \mu_2)}$

The change in path difference due to the two slabs is $(\mu_1 – \mu_2)t$ (as in problem no. 16). $\\$ For having a minimum at $P_0$, the path difference should change by $\lambda/2$.

18   A thin paper of thickness $0\cdot02\ mm$ having a refractive index $1\cdot45$ is pasted across one of the slits in a Young's double slit experiment. The paper transmits $4/9$ of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also ? The wavelength of the light used is $600\ nm$.

##### Solution :

where, $r_1$ and $r_2$ are corresponding amplitudes.

a) Let, $I_1$ = Intensity of source without paper = $I$ $\\$ b) Then $I_2$ = Intensity of source with paper = $(4/9)I$

b) No. of fringes that will cross the origin is given by, $\\$ $n = \dfrac{(\mu - 1)t}{\lambda} = \dfrac{(1.45 -1) \times 0.02 \times 10^{-3}}{600 \times 10^{-9}} = 15$

$\Rightarrow \dfrac{I-1}{I-2} = \dfrac{9}{4} \Rightarrow \dfrac{r_1}{r_2} = \dfrac{3}{2}\ [\ because\ I\ \infty\ r^2\ ]$

Given that, $t = 0.02\ mm = 0.02 \times 10^{–3} m,\ \mu_1 = 1.45,\ \lambda = 600\ nm = 600 \times 10^{–9}\ m$

So, $\dfrac{I_{max}}{I_{min}} = \dfrac{(r_1 + r_2)^2}{(r_1 - r_2)^2} = 35 : 1$

$\Rightarrow \dfrac{I_1}{I_2} = \dfrac{9}{4} \Rightarrow \dfrac{r_1}{r_2} = \dfrac{3}{2}\ [\ because\ I\ \infty\ r^2\ ]$

19   A Young's double slit apparatus has slits separated by $0\cdot28\ mm$ and a screen $48\ cm$ away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light ($\lambda = 700\ nm$ in vacuum). Find the fringe-width of the pattern formed on the screen.

##### Solution :

Let, $\lambda_w$ = wavelength of red light in water Since, the fringe width of the pattern is given by,

Given that, $d = 0.28\ mm = 0.28 \times 10^{–3}\ m, D = 48\ cm = 0.48\ m,\ \lambda_a = 700\ nm$ in vacuum

$\beta = \dfrac{\lambda_{w}D}{d} = \dfrac{525 \times 10^{-9}\times 0.48}{0.28 \times 10^{-3}} = 9 \times 10^{-4}\ m = 0.90\ mm.$

20   A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle $\theta$ = $sin^{-1}$ $\big[\frac{\lambda}{2d}\big]$ with the normal to the plane of 2d the slits, there will be a dark fringe at the centre $P_{0}$ of the pattern.

##### Solution :

In the $\Delta S_1S_2X$, $\\$ $sin\ \theta = \dfrac{S_2X}{S_1S_2}$

As the path difference is an odd multiple of $\lambda/2$, there will be a dark fringe at point $P_0$

It can be seen from the figure that the wavefronts reaching $O$ from $S_1$ and $S_2$ will have a path difference of $S_2X$.

So path diffrences $= S_2X = S_1S_2\ sin\theta = d\ sin\theta = d \times \dfrac{\lambda}{2d} = \dfrac{\lambda}{2}$

21   A narrow slit $S$ transmitting light of wavelength $\lambda$ is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen $\sum$ placed at a distance $D$ from the slit. (a) What will be the intensity at a point just above the mirror, i.e., just above $O$ ? (b) At what distance from $O$ does the first maximum occur ?

##### Solution :

$\Rightarrow \dfrac{y \times 2d}{D} = n{\lambda} - \frac{\lambda}{2} \Rightarrow y = \frac {\lambda{D}}{4d}$

b) Here, $2d$ = equivalent slit separation $D$ = Distance between slit and screen. $\\$ We know for bright fringe, $\Delta{x} = \dfrac{y\times 2d}{D} = n\lambda$

$\Rightarrow \dfrac{y \times 2d}{D} + \frac{\lambda}{2} = n{\lambda}$

a) Since, there is a phase difference of $\pi$ between direct light and reflecting light, the intensity just above the mirror will be zero.

But as there is a phase reversal of $\lambda/2$ .

22   A long narrow horizontal slit is placed $1\ mm$ above a horizontal plane mirror. The interference between the light coming directly from the slit and that after reflection is seen on a screen $1\cdot0\ m$ away from the slit. Find the fringe-width if the light used has a wavelength of $700\ nm$.

##### Solution :

Fringe width $= \dfrac{\lambda{D}}{d} = \dfrac{700 \times 10^{-9}\times 1m}{2\times 10^{-3}\ m} = 0.35\ mm$.

Given that, $D = 1\ m, \lambda = 700\ nm = 700 \times 10^{–9}\ m$ Since, $a = 2\ mm, d = 2a = 2\ mm = 2 \times 10^{–3}\ m$ (L loyd’s mirror experiment)

23   Consider the situation of the previous problem. If the mirror reflects only $64$% of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen ?

##### Solution :

Given that, the mirror reflects 64% of energy (intensity) of the light. $\\$ So, $\dfrac{I_1}{I_2} = 0.64 = \dfrac{16}{25} \Rightarrow \dfrac{r_1}{r_2} = \dfrac{4}{5}$ $\\$ So, $\dfrac{I_{max}}{I_{min}} = \dfrac{(r_1 + r_2)^2}{(r_1 - r_2)} = 81 : 1$

24   A double slit $S_{1} — S_{2}$ is illuminated by a coherent light of wavelength $\lambda$. The slits are separated by a distance $d$. A plane mirror is placed in front of the double slit at a distance $D_{1}$ from it and a screen $\sum$ is placed behind the double slit at a distance $D_{2}$ from it (figure 17-E2). The screen $\sum$ receives only the light reflected by the mirror. Find the fringe-width of the interference pattern on the screen.

##### Solution :

It can be seen from the figure that, the apparent distance of the screen from the slits is, $\\$ $D = 2D_1 + D_2$ $\\$ So, Fringe width $= \dfrac{D\lambda}{d} = \dfrac{(2D_1 + D_2)\lambda}{d}$

25   White coherent light ($400\ nm\ -700\ nm$) is sent through the slits of a Young's double slit experiment (figure 17-E3). The separation between the slits is $0\cdot5\ mm$ and the screen is $50\ cm$ away from the slits. There is a hole in the screen at a point $1\cdot0\ mm$ away (along the width of the fringes) from the central line. (a) Which. wavelength(s) will be absent in the light coming from the hole ? (b) which wavelength(s) will have a strong intensity ?

##### Solution :

Again, where $n = 2, \lambda_2 = \dfrac{Y_n d}{2D} = 500\ nm$

If $n = 1, \lambda_1 = \big(\frac{2}{3}\big) \times 1000 = 667\ nm$ $\\$ If $n = 1, \lambda2 = \big(\frac{2}{5}) \times 1000 = 400\ nm$

$y_n = \dfrac{n\lambda_n D}{d} \Rightarrow \lambda_n = \dfrac{Y_n d}{nD}$

$D = 50\ cm = 0.5\ m$ and on the screen $y_n = 1\ mm = 1 \times 10^{–3}\ m$

b) For strong intensity (bright fringes) at the hole

$1000\ nm$ is not present in the range $400\ nm\ –\ 700\ nm$

$\Rightarrow \lambda_n = \dfrac{2}{2n +1}\dfrac{\lambda_{n}d}{D} = \dfrac{2}{2n +1} \times \dfrac{10^{-3} \times 0.5 \times 10^{-3}}{0.5} \Rightarrow \dfrac{2}{2n +1} \times 10^{-6}\ m = \dfrac{2}{2n +1}\times 10^{3}\ nm$

Given that, $\lambda = (400\ nm\ to\ 700\ nm),\ d = 0.5\ mm = 0.5 \times 10^{–3}\ m$,

So, the only wavelength which will have strong intensity is $500\ nm$.

So, the light waves of wavelengths $400\ nm$ and $667\ nm$ will be absent from the out coming light.

When, $n = 1, \lambda_1 = \dfrac{Y_n d}{nD} = \dfrac{10^{-3}\times 0.5\times 10^{-3}}{0.5} = 10^{-6}\ m =- 1000\ nm.$

a) We know that for zero intensity (dark fringe) $\\$ $y_n = \big(\dfrac{2n +1}{2}\big)\dfrac{\lambda_{n}D}{d}$ where $n = 0, 1, 2, .......$

26   Consider the arrangement shown in figure (17-E4). The distance $D$ is large compared to the separation d between the slits. (a) Find the minimum value of $d$ so that there is a dark fringe at $O$. (b) Suppose $d$ has this value. Find the distance $x$ at which the next bright fringe is formed. (c) Find the fringe-width.

##### Solution :

So, $2(\sqrt{d^2 + D^2} -D) = (2n + 1)\big(\frac{\lambda}{2}\big)$

Path difference $= (AB + BO)\ –\ (AC + CO)$

For dark fringe, path difference should be odd multiple of $\lambda/2$.

From the diagram, it can be seen that at point $O$.

$2(AB\ –\ AC)$ [Since, $AB = BO$ and $AC = CO$] = $2(\sqrt{d^2 + D^2} - D)$

27   Two coherent point sources $S_{1}$ and $S_{2}$ vibrating in phase emit light of wavelength $\lambda$. The separation between the sources is $2\lambda$. Consider a line passing through $S_2$ and perpendicular to the line $S_1\ S_2$. What is the smallest distance from $S_2$ where a minimum of intensity occurs ?

##### Solution :

$n = 1 \Rightarrow a = \frac{7\lambda}{12} \ \ \ \ \ \ \ n = 2 \Rightarrow z = \frac{-9\lambda}{20}$

For minimum intensity $\\$ $\therefore S_1P – S_2P = x = (2n +1) \lambda/2$ $\\$ From the figure, we get $\\$ $\Rightarrow \sqrt{z^2 + (2\lambda)^2} -z = (2n + 1)\dfrac{\lambda}{2}$ $\\$ $\Rightarrow z^2 + 4\lambda = z^2 + 9 (2n +1)^2\dfrac{\lambda^2}{4} + z(2n +1)\lambda$ $\\$ $\Rightarrow z = \dfrac{4\lambda^2 - (2n + 1)^2(\lambda^2/4)}{(2n + 1)\lambda} = \dfrac{16\lambda^2-(2n +1)^2\lambda^2}{4(2n + 1)\lambda} \ \ \ \ \ \ \ .....(1)$ $\\$ putting , $n = 0 \Rightarrow Z = \frac{15\lambda}{4} \ \ \ \ \ \ \ \ \ n = -1\Rightarrow z = \frac{15\lambda}{4}$ $\\$

$\therefore Z = 7\lambda/12$ is the smallest distance for which there will be minimum intensity.

28   Figure (17-E5) shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $BP_0 — AP_0 = \lambda/ 3$ and $D>> \lambda$. (a) Show that in this case $d$ = $\sqrt{2\lambda{D}/3}$ . (b) Show that the intensity at $P_0$ is three times the intensity due to any of the three slits individually.

##### Solution :

$\therefore$ path difference $= \Delta{x} = n\lambda$ $\\$ From the figure, $\\$

Since $S_1,\ S_2$ are in same phase, at $O$ there will be maximum intensity. Given that, there will be a maximum intensity at $P$.

$(S_1P)^2 - (S_2P)^2 = (\sqrt{D^2 + X^2})^2 - (\sqrt{(D - 2\lambda) + X^2})^2$ $\\$ $= 4\lambda{D} - 4\lambda^2 = 4\ \lambda{D}$ ($\lambda^2$ is so small and can be neglected) $\\$ $\Rightarrow S_1P - S_2P = \dfrac{4\lambda{D}}{2\sqrt{X^2 + D^2}} = n\lambda$ $\\$ $\Rightarrow \dfrac{2D}{\sqrt{X^2 + D^2}} = v$ $\\$ $\Rightarrow n^2 (X^2 +D^2) = 4D^2 = \Delta{X} = \frac{D}{n}\sqrt{4 - n^2}$ $\\$ when $n = 1,\ x = \sqrt{3D} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1^{st}\ order)$ $\\$ $n = 2,\ x = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2^{nd}\ order)$ $\\$ $\therefore$ When $X = \sqrt{3D}$, at $P$ there will be maximum intensity.

29   In a Young's double slit experiment, the separation between the slits = $2\cdot0\ mm$, the wavelength of the light = $600\ nm$ and the distance of the screen from the slits = $2\cdot0\ m$. If the intensity at the centre of the central maximum is $0\cdot20\ W/m^2$, what will be the intensity at a point $0\cdot5\ cm$ away from this centre along the width of the fringes ?

##### Solution :

As shown in the figure, $\\$ $(S_1P)^2 = (PX)^2 + (S_1X)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......(1)$ $\\$ $(S_2P)^2 = (PX)^2 + (S_2X)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ .......(2)$ $\\$ From (1) and (2), $\\$ $(S_1P)^2 - (S_2P)^2 = (S_1X)^2 - (S_2X)^2$ $\\$ $= (1.5\lambda + R\ cos\ \theta)^2 - ( R\ cos\ \theta - 15\lambda)^2$ $\\$ $= 6\lambda R\ cos\ \theta$ $\\$ $(S_1P - S_2P) = \dfrac{6\lambda R\ cos\ \theta}{2R} = 3\lambda\ cos\ \theta$ . $\\$ For constructive interference, $\\$ $(S_1P - S_2P)^2 = x = 3\lambda\ cos\ \theta = n\lambda$ $\\$ $\Rightarrow cos\theta = \dfrac{n}{3} \Rightarrow \theta = cos^{-1}\big(\dfrac{n}{3}\big) , where n = 0,\ 1,\ 2,\ .....$ $\\$ $\Rightarrow \theta = 0^o,\ 48.2^o,\ 70.5^o,\ 90^o,$ and similar points in other quadrants.

30   In a Young's double slit interference experiment the fringe pattern is observed on a screen placed at a distance $D$ from the slits. The slits are separated by a distance $d$ and are illuminated by monochromatic light of wavelength $\lambda$. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

##### Solution :

$\Rightarrow d = \sqrt{(2\lambda{D})/3}$ (neglecting the term $\lambda^2/9$ as it is very small). $\\$ b) To find the intensity at $P_0$, we have to consider the interference of light waves coming from all the three slits.$\\$ Here, $CP_0\ –\ AP_0 = \sqrt{D^2 + 4d^2} - D$ $\\$ = $\sqrt{D^2 + \dfrac{8\lambda{D}}{3}} - D$ = $D \big(1 + \dfrac{8\lambda}{3D}\big)^\frac{1}{2} - D$

Again, $\phi_b = \frac{2\pi{x}}{3\lambda} = \frac{2\pi}{3} \ \ \ \ \ \ \ \ \ \ \ .....(2)$ $\\$ So, it can be said that light from B and C are in same phase as they have some phase difference with respect to A. So, $R = \sqrt{(2r)^2 + r^2 + 2 \times 2r \times r\ cos(\frac{2\pi}{3})} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (using\ vector\ method)$ $\\$ $= \sqrt{4r^2 + r^2 - 2r^2} = \sqrt{3r}$ $\\$ $\therefore I_{P_O} - K (\sqrt{3r})^2 = 3Kr^2 = 3I$ $\\$ As, the resulting amplitude is $\sqrt{3}$ times, the intensity will be three times the intensity due to individual slits.

a) As shown in the figure, $BP_0\ –\ AP_0 = \dfrac{\lambda}{3}$ $\\$ $\Rightarrow \sqrt{(D^2 + d^2 )} - D =\dfrac{\lambda}{3}$ $\\$ $\Rightarrow D^2 + d^2 = D^2 + \big(\dfrac{\lambda^2}{9}\big) + \dfrac{(2\lambda{D})}{3}$

$= D \big(1 + \frac{8\lambda}{3D \times 2} + .......\big) - D = \dfrac{4\lambda}{3}$ [using binomial expansion] $\\$ So, the corresponding phase difference between waves from C and A is, $\\$ $\phi_c = \dfrac{2\pi{x}}{\lambda} = \dfrac{2\pi \times 4\lambda}{3\lambda} = \dfrac{8\pi}{3} = \big( 2\pi + \dfrac{2\pi}{3}\big) = \dfrac{2\pi}{3} \ \ \ \ \ \ \ \ \ \ \ \ .......(1)$

31   In a Young's double slit experiment $\lambda = 500\ nm, d = 1\cdot0 mn$ and $D = 1\cdot0\ m$. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

##### Solution :

So, the amplitude of the resulting wave at the point y = 0.5 cm is, $\\$ $A = \sqrt{r^2 + r^2 + 2r^2 cos(\frac{2\pi}{3})} = \sqrt{r^2 + r^2 - r^2} = r$ $\\$ Since, $\dfrac{I}{I _{max}} = \dfrac{A^2}{2r^2}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$[since, maximum amplitude = $2r$]

Given that, $d = 2\ mm = 2 \times 10^{–3}\ m, \lambda = 600\ nm = 6 \times 10^{–7}\ m, I_{max} = 0.20 W/m^2, D = 2\ m$ $\\$ For the point, $y = 0.5\ cm$ $\\$ We know, path difference $= x = \dfrac{yd}{D} = \frac{0.5 \times 10^{-2} \times 2 \times 10^{-3}}{2} = 5 \times 10^{-6}\ m$ $\\$ So, the corresponding phase difference is, $\\$ $\phi = \dfrac{2\pi{x}}{\lambda} = \dfrac{2\pi \times 5 \times 10{-6}}{6 \times 10^{-7}} \Rightarrow \dfrac{50\pi}{3} = 16\pi + \dfrac{2\pi}{3}$

$\Rightarrow \dfrac{I}{0.2} = \dfrac{A^2}{4r^2} = \dfrac{r^2}{4r^2}$ $\\$ $\Rightarrow I = \dfrac{0.2}{4} = 0.05 W/m^2$

32   The linewidth of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the linewidth of a bright fringe in a Young's double slit experiment in terms of $\lambda$, $d$ and $D$ where the symbols have their usual meanings.

##### Solution :

$\Rightarrow$ Path difference, $x =\dfrac{\lambda}{4}$ $\\$ $\Rightarrow y = \dfrac{xD}{d} =\dfrac{\lambda{D}}{4d}$ $\\$ ii) When intensity is $1/4^th$ of the maximum $\dfrac{I}{I_{max}} = \dfrac{1}{4}$ $\\$ $\Rightarrow \dfrac{4a^2\ cos^2(\phi/2)}{4a^2} = \dfrac{1}{4}$ $\\$ $cos^2(\phi/2) = \dfrac{1}{4} \Rightarrow cos(\phi/2) = \frac{1}{2}$ $\\$ $\Rightarrow \dfrac{\phi}{2} = \dfrac{\pi}{4} \Rightarrow \phi = \dfrac{2\pi}{3}$

When intensity is half the maximum $\dfrac{I}{I_{max}} = \dfrac{1}{2}$ $\\$ $\Rightarrow \dfrac{4a^2\ cos^2(\phi/2)}{4a^2} = \dfrac{1}{2}$ $\\$ $cos^2(\phi/2) = \dfrac{1}{2} \Rightarrow cos(\phi/2) = \frac{1}{\sqrt{2}}$ $\\$ $\Rightarrow \dfrac{\phi}{2} = \dfrac{\pi}{4} \Rightarrow \phi = \dfrac{\pi}{2}$

$\Rightarrow$ Path difference, $x =\dfrac{\lambda}{3}$ $\\$ $\Rightarrow y = \dfrac{xD}{d} =\dfrac{\lambda{D}}{3d}$

33   Consider the situation shown in figure (17-E6). The two slits $S_1$ and $S_2$ p laced symmetrically around the central line are illuminated by a monochromatic light of wavelength $\lambda$. The separation between the slits is $d$. The light transmitted by the slits falls on a screen $\sum_1$ placed at a distance $D$ from the slits. The slit $S_3$ is at the central line and the slit $S_4$ is at a distance $z$ from $S_3$. Another screen $\sum_2$ is placed a further distance $D$ away from $\sum_{1}$,. Find the ratio of the maximum to minimum intensity observed on $\sum_2$, if $z$ is equal to $\\$ (a) $z = \frac{\lambda{D}}{2d}$ $\\$ (b) $\frac{\lambda{D}}{d}$ $\\$ (c) $\frac{\lambda{D}}{4d}$

##### Solution :

Given that, $D = 1\ m, d = 1\ mm = 10^{–3}\ m, \lambda = 500\ nm = 5 \times 10^{–7}\ m$ $\\$ For intensity to be half the maximum intensity. $\\$ $y = \dfrac{\lambda{D}}{4d}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (As in problem no. 32) $\\$ $\Rightarrow y = \dfrac{5 \times 10^{-7} \times 1}{4 \times 10^{-3}} \Rightarrow y = 1.25 \times 10^{-4}\ m.$

34   Consider the arrangement shown in figure (17-E7). By some mechanism, the separation between the slits $S_3$ and $S_4$ can be changed. The intensity is measured at the point $P$ which is at the common perpendicular bisector of $S_1\ S_2$ and $S_3\ S_4$. When $z = \frac{D\lambda}{2d}$, the intensity measured at $P$ is $I$. Find this intensity when $z$ is equal to $\\$ (a) $\frac{D\lambda}{d}$ $\\$ (b) $\frac{3D\lambda}{2d}$ , and $\\$ (c) $\frac{2D\lambda}{d}$

##### Solution :

The line width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. We know that, for intensity to be half the maximum $\\$ $y = \pm\dfrac{\lambda{D}}{4d}$ $\\$ $\therefore$ Line width $= \dfrac{\lambda{D}}{4d} + \dfrac{\lambda{D}}{4d} = \dfrac{\lambda{D}}{2d}$

35   A soap film of thickness $0\cdot0011\ mm$ appears dark when seen by the reflected light of wavelength $580\ nm$. What is the index of refraction of the soap solution, if it is known to be between $1\cdot2$ and $1\cdot5$ ?

##### Solution :

i) When, $z = \dfrac{\lambda{D}}{2d}$, at $S_4$, minimum intensity occurs (dark fringe) $\\$ $\Rightarrow$ Amplitude = $0$, $\\$ At $S_3$, path difference = $0$ $\\$ $\Rightarrow$ Maximum intensity occurs. $\\$ $\Rightarrow$ Amplitude = $2r$. $\\$ So, on $\sum{2}$ screen, $\\$ $\dfrac{I_{max}}{I_{min}} = \dfrac{(2r + 0)^2}{(2r + 0)^2} = \infty$ $\\$ ii) When, $z = \dfrac{\lambda{D}}{2d}$, At $S_4$, minimum intensity occurs. (dark fringe) $\\$ $\Rightarrow$ Amplitude = $0$.$\\$ At $S_3$, path difference = $0$ $\\$ $\Rightarrow$ Maximum intensity occurs. $\\$ $\Rightarrow$ Amplitude = $2r$. $\\$ So, on $\sum{2}$ screen, $\\$ $\dfrac{I_{max}}{I_{min}} = \dfrac{(2r + 2r)^2}{(2r - 0)^2} = 1$ $\\$ iii) When, $z = \dfrac{\lambda{D}}{4d},$ At $S_4$, intensity $= \dfrac{I_{max}}{2}$ $\\$ $\Rightarrow$ Amplitude = $\sqrt{2r}$ . $\\$ $\Rightarrow$ At $S_3$, intensity is maximum. $\\$ $\Rightarrow$ Amplitude = $2r.$ $\\$ $\dfrac{I_{max}}{I_{min}} = \dfrac{(2r + \sqrt{2r})}{(2r - \sqrt{2r})} = 34$.

36   A parallel beam of light of wavelength $560\ nm$ falls on a thin film of oil (refractive index = $1\cdot4$). What should be the minimum thickness of the film so that it strongly reflects the light ?

##### Solution :

$\dfrac{I"}{I} = \dfrac{a^2 + a^2 + 2a^2 cos\big(\dfrac{3\pi}{2})}{a^2 + a^2 + 2a^2 cos \dfrac{\pi}{2}} = \dfrac{2a^2}{2a^2} = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow I'' = I' =\dfrac{I}{4}$

$\dfrac{I"}{I} = \dfrac{a^2 + a^2 + 2a^2 cos 2\pi}{a^2 + a^2 + 2a^2 cos \dfrac{\pi}{2}} = \dfrac{4a^2}{2a^2} = 2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow I'''' = 2I' = 2\dfrac{I}{4} = \dfrac{I}{2}$

$\Rightarrow$ At $S_3$, intensity at $S_3 = 0 \Rightarrow I_1 = 0$ $\\$ At $S_4$, intensity at $S_4 = 0 \Rightarrow I_2 = 0$ $\\$ At $P$, path difference $= 0 \Rightarrow$ Phase difference = $0$. $\\$

c) When $z = \dfrac{2D}{d}$ $\\$ $\Rightarrow y = OS_3 = OS_4 = \dfrac{D\lambda}{d}$ $\\$ $\therefore \phi = \dfrac{2\pi{x}}{\lambda} = \dfrac{2\pi}{\lambda} \times \dfrac{yd}{D} = \dfrac{2\pi}{\lambda} \times \dfrac{D\lambda}{d} \times \dfrac{d}{D} = 2\pi$ $\\$ Let, $I''''$ = intensity at $S_3$ and $S_4$ when, $\phi = 2\pi.$

When, $z = \dfrac{3D}{2d}, \Rightarrow y = \dfrac{3D\lambda}{4d}$ $\\$ $\therefore \phi = \dfrac{2\pi{x}}{\lambda} = \dfrac{2\pi}{\lambda} \times \dfrac{yd}{D} = \dfrac{2\pi}{\lambda} \times \dfrac{3D\lambda}{4d} \times \dfrac{d}{D} = \dfrac{3\pi}{2}$ $\\$ Let, I" be the intensity at $S_3$ and $S_4$ when, $\phi = \dfrac{3\pi}{2}$ Now comparing,

a) When, $z = \dfrac{D\lambda}{d}$ $\\$ So, $OS_3 = OS_4 = \dfrac{D\lambda}{2d} \Rightarrow$ Dark fringe at $S_3$ and $S_4$.

$\therefore$ Intensity at $P = \dfrac{I}{4} + \dfrac{I}{4} + 2 \times \big(\dfrac{I}{4}\big) cos 0° = \dfrac{I}{2} + \dfrac{I}{2} = I$.

At $P,\ I_{resultant} = \dfrac{I}{2} + \dfrac{I}{2} + 2\big(\dfrac{I}{2}\big) cos 0° = I + I = 2I$. So, the resultant intensity at $P$ will be $2I$.

$\Rightarrow I = I_1 + I_2 + \sqrt{I_1I_2} cos 0° = 0 + 0 + 0 = 0 \Rightarrow$ Intensity at $P = 0$. $\\$ b) Given that, when $z = {D\lambda}{2d}$, intensity at $P = I$ $\\$ Here,$OS_3 = OS_4 = Y = \dfrac{D\lambda}{4d}$ $\\$ $\therefore \phi = \dfrac{2\pi{x}}{\lambda} = \dfrac{2\pi}{\lambda} \times \dfrac{yd}{D} = \dfrac{2\pi}{\lambda} \times \dfrac{D\lambda}{4d} \times \dfrac{d}{D} = \dfrac{\pi}{2}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ [Since, $x$ = path difference = $yd/D$] $\\$ Let, intensity at $S_3$ and $S_4 = I'$ $\\$ $\therefore$ At $P$, phase difference = $0$ $\\$ So, $I' + I' + 2I' cos 0° = I$. $\\$ $\Rightarrow 4I' = I \Rightarrow I' = \dfrac{1}{4}$.

37   A parallel beam of white light is incident normally on a water film $1\cdot0 \times 10^{-4}\ cm$ thick. Find the wavelength in the visible range ($400\ nm - 700\ nm$) which are strongly transmitted by the film. Refractive index of water = $1\cdot33$.

##### Solution :

$\Rightarrow \mu = \dfrac{n\lambda}{2d} = \dfrac{2n\lambda}{4d} = \dfrac{580 \times 10^{-9} \times 2n}{4 \times 11 \times 10^{-7}} = \dfrac{5.8}{44}(2n) = 0.132(2n)$ $\\$ Given that, $\mu$ has a value in between $1.2$ and $1.5$. $\\$ $\Rightarrow$ When, $n = 5,\ \mu = 0.132 \times 10 = 1.32.$

Given, $d = 0.0011 \times 10^{–3}\ m$ $\\$ For minimum reflection of light, $2\mu{d} = n\lambda$

38   A glass surface is coated by an oil film of uniform thickness $1\cdot00 \times 10^{-4}\ cm$. The index of refraction of the oil is $1\cdot25$ and that of the glass is $1\cdot50$. Find the wavelengths of light in the visible region ($400\ nm - 750\ nm$) which are completely transmitted by the oil film under normal incidence.

##### Solution :

Given that, $\lambda= 560 \times 10^{–9}\ m,\ \mu = 1.4.$ $\\$ For strong reflection, $2\mu{d} = (2n + 1)\dfrac{\lambda}{2} \Rightarrow d = \dfrac{(2n 1)\lambda}{4d}$ $\\$ For minimum thickness, putting $n = 0.$ $\\$ $\Rightarrow d = \dfrac{\lambda}{4d} \Rightarrow d = \dfrac{560 \times 10^{-9}}{14} = 10^{–7}\ m = 100\ nm$

39   Plane microwaves are incident on a long slit having a width of $5\cdot0\ cm$. Calculate the wavelength of the microwaves if the first diffraction minimum is formed at $\theta = 30°$.

##### Solution :

For strong transmission, $2 \mu{d} = n\lambda \Rightarrow \lambda = \dfrac{2\mu{d}}{n}$ $\\$ Given that, $\mu = 1.33, d = 1 \times 10^{–4}\ cm = 1 \times 10^{–6}\ m$. $\\$ $\Rightarrow \lambda =\dfrac{2 \times 1.33 \times 1 \times 10^{-6}}{n} = \dfrac{2660}{10^{-9}}\ m$ $\\$ when, $\ \ n = 4, \lambda{1} = 665\ nm$ $\\$ $\ \ \ \ \ \ \ \ \ \ \ n = 5, \lambda{2} = 532\ nm$ $\\$ $\ \ \ \ \ \ \ \ \ \ \ n = 6, \lambda{3} = 443\ nm$ $\\$

40   Light of wavelength $560\ nm$ goes through a pinhole of diameter $0\cdot20\ mm$ and falls on a wall at a distance of $2\cdot00\ m$. What will be the radius of the central bright spot formed on the wall ?

##### Solution :

$\Rightarrow \lambda = \dfrac{5000\ nm}{2n + 1}$ $\\$ For the wavelengths in the region $(400\ nm\ –\ 750\ nm)$ $\\$ When, $n = 3, \lambda = \dfrac{5000}{2 \times 3 + 1} = \dfrac{5000}{7} = 714.3\ nm$

For the thin oil film, $\\$ $d = 1 \times 10^{–4}\ cm = 10^{–6}\ m,\ \mu_{oil} = 1.25$ and $\mu_{x} = 1.50$ $\\$ $\lambda = \dfrac{2\mu{d}}{(n+ 1/ 2)} \dfrac{2 \times 10^{-6} \times 1.25 \times 2 }{ 2n + 1} = \dfrac{5 \times 10^{-6}\ m}{2n + 1}$

When, $n = 4, \lambda = \dfrac{5000}{2 \times 4 + 1} = \dfrac{5000}{9} = 555.6\ nm$ $\\$ When, $n = 5, \lambda = \dfrac{5000}{2 \times 5 + 1} = \dfrac{5000}{11} = 454.5\ nm$

41   A convex lens of diameter $8\cdot0\ cm$ is used to focus a parallel beam of light of wavelength $620\ nm$. If the light be focused at a distance of $20\ cm$ from the lens, what would be the radius of the central bright spot formed ?

##### Solution :

For first minimum diffraction, $b\ sin\ \theta = \lambda$ $\\$ Here, $\theta = 30°,\ b = 5\ cm$ $\\$ $\therefore \lambda = 5 \times sin 30° = \dfrac{5}{2} = 2.5\ cm.$