Concept Of Physics Light Waves

H C Verma

Concept Of Physics

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

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

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

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

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.

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

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

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

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

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

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.

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

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

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

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

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.

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

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.

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

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

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

b) When, $\beta = 1\ mm = 10^{–3}\ m$

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

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.

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

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

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 ?

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 ?

Answer

8   None

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,

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 $2\cdot0\ \times 10^{-3}\ M$.

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

As shown in the figure, angular separation