Ray Optics and Optical Instruments

Physics

NCERT

1   A small candle, $2.5 cm$ in size is placed at $27 cm$ in front of a concave mirror of radius of curvature $36 cm$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Solution :

Size of the candle,$ h = 2.5 cm$$\\$ Image size = $h’$$\\$ Object distance, $u = - 27 cm$$\\$ Radius of curvature of the concave mirror,$ R = -36 cm$$\\$ Focal length of the concave mirror,$f=\dfrac{R}{2}=-18cm$$\\$ $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}\\ \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\ =\dfrac{1}{-18}-\dfrac{1}{-27}\\ =\dfrac{-3+2}{54}=-\dfrac{1}{54}\\ \therefore V=-54 cm $$\\$ Therefore, the screen should be placed $54 cm$ away from the mirror to obtain a sharp image. The magnification of the image is given as:$\\$ $m=\dfrac{h'}{h}=\dfrac{v}{u}\\ \therefore h'=-\dfrac{v}{u}*h\\ =-(\dfrac{-54}{-27})*2.5\\ =-5 cm$$\\$ The height of the candle’s image is $ 5 cm$ . The negative sign indicated that the image is inverted and real.$\\$ If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

2   A $4.5 cm$ needle is placed $12 cm$ away from a convex mirror of focal length $15 cm.$ Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Solution :

Height of the needle, $h_1 = 4.5 cm$$\\$ Object distance, $u = -12 cm$$\\$ Focal length of the convex mirror, $f = 15 cm$$\\$ Image distance = $v$ The value of v can be obtained using the mirror formula:$\\$ $\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}\\ \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\ =\dfrac{1}{15}+\dfrac{1}{12}=\dfrac{4+5}{60}\\ =\dfrac{9}{60}\\ \therefore v=\dfrac{60}{9}=6.7 cm$$\\$ Hence, the image of the needle is $6.7 cm$ away from the mirror. Also, it is on the other side of the mirror.$\\$ The image size is given by the magnification formula:$\\$ $m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}\\ \therefore h_2=-\dfrac{v}{u}*h_1\\ =\dfrac{-67}{-12}*4.5=+2.5 cm$$\\$ Hence, Magnification of the image,$m=\dfrac{h_2}{h_1}\\ =\dfrac{2.5}{4.5}=0.56 $$\\$ The height of the image is $2.5 cm$. The positive sign indicated that the image is erect, Virtual, and diminished. If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

3   A tank is filled with water to a height of $12.5 cm$. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be $9.4 cm.$ What is the refractive index of water? If water is replaced by a liquid of refractive index $1.63$ up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Solution :

Actual depth of the needle in water, $h_1 =12.5 cm$$\\$ Apparent depth of the needle in water, $h_2 =9.4 cm$$\\$ Refractive index of water $=\mu$$\\$ The value of $\mu $ can be obtained as follows:$\\$ $\mu =\dfrac{h_1}{h_2}\\ =\dfrac{12.5}{9.4}\approx 1.33$$\\$ Hence, the refractive index of water is about $1.33$ Water is replaced by a liquid of refractive index, $\mu ' = 1.63$$\\$ The actual depth of the needle remains the same, but its apparent depth changes.$\\$ Let y be the new apparent depth of the needle. Hence, we can write the relation:$\\$ $\mu'=\dfrac{h_1}{y}\\ \therefore y=\dfrac{h_1}{\mu'}\\ =\dfrac{12.5}{1.63}\\ =7.67 cm$$\\$ Hence, the new apparent depth of the needle is$7.67 cm$. It is less than $h_2 $. Therefore, to focus the needle again, the microscope should be moved up. Distance by which the microscope should be moved up $= 9.4 - 7.67 = 1.73 cm$

4   Figures (a) and (b) show refraction of a ray in air incident at with $60^o$ with the normal to a glass-air and water air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is $45^o$ with the normal to a water-glass interface [figure (c)]

Solution :

As per the given figure, for the glass -air interface:$\\$ Angle of incidence, $I = 60^o$$\\$ Angle of refraction, $r = 35^o$$\\$ The relative refractive index of glass with respect to air is given by Snell’s law as:$\\$ $\mu_g^a=\dfrac{\sin i}{\sin r}\\ =\dfrac{\sin 60^o}{\sin 35^o}=\dfrac{0.8660}{0.5736}=1.51 $$\\$ As per the given figure, for the air-water interface:$\\$ Angle of incidence, $i = 60^o$$\\$ Angle of refraction, $r = 47^o$$\\$ The relative refractive index of water with respect to air is given by Snell’s law as:$\\$ $\mu_w^a=\dfrac{\sin i}{\sin r}\\ \dfrac{\sin 60}{\sin 47}=\dfrac{0.8660}{0.7314}\\ =1.184$$\\$ Using (1) and (2), the relative refractive index of glass with respect to Water can be Obtained as:$\\$ $\mu_g^w=\dfrac{\mu_g^a}{\mu_w^a}\\ =\dfrac{1.51}{1.184}=1.275$$\\$ The following figure shows the situation involving the glass - water interface. $\\$ Angle if incidence, $i = 45 ^o$$\\$ Angle of refraction $= r$ From Snell’s law,$ r$ can be calculated as:$\\$ $\dfrac{\sin i}{\sin r}=\mu_g^w\\ \dfrac{\sin 45^o}{\sin r}=1.275\\ \sin r =\dfrac{\dfrac{1}{\sqrt{2}}}{1.275}=0.5546\\ \therefore r=\sin^{-1}(0.5546)=38.68^o$

5   A small bulb is placed at the bottom of a tank containing water to a depth of $80 cm$. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33.$ (Consider the bulb to be a point source.)

Solution :

Actual depth of the bulb in water, $d_1 = 80 cm = 0.8 m$$\\$ Refractive index of water, $\mu = 1.33$$\\$ The given situation is shown in the following figure:$\\$ Where,$\\$ $i =$ Angle of incidence$\\$ $r =$ Angle of refraction $= 90^o$$\\$ Since the bulb is a point source, the emergent light can be considered as a circle of

Radius, $R =\dfrac{AC}{2}=AO=OB$$\\$ Using Snell’s Law, we can write the relation for the refractive index of water as:$\\$ $\mu =\dfrac{\sin r}{\sin i}\\ 1.33=\dfrac{\sin 90^o}{\sin i}\\ \therefore i=\sin^{-1}(\dfrac{1}{1.33})=48.75^o$$\\$ Using the given figure, we have the relation:$\\$ $\tan i =\dfrac{OC}{OB}=\dfrac{R}{d_1}$$\\$ $R=\tan 48.75^o*0.8=0.91m$$\\$ Area of the surface of water$=\pi R^2=\pi(0.91)^2=2.61m^2$$\\$ Hence, the area of the surface of water through which the light from the bulb can emerge is approximately $2.61 m^2 $.