# 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?

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$$\\ Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m^2 . 6 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40^o . What is the refractive index of the material of the prism? The refracting angle of the prism is 60^o . If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. ##### Solution : Angle of minimum deviation, \delta m = 40^o$$\\$ Angle of the prism, $A = 60 ^o$$\\ Refractive index of water, \mu = 1.33$$\\$ Refractive index of the material of the prism = $\mu '$$\\ The angel of deviation is related to refractive index (\mu') as:\\ \mu=\dfrac{\sin(\dfrac{A+\delta_m}{2})}{\sin\dfrac{A}{2}}\\ =\dfrac{\sin\dfrac{(60^o+40)^o}{2}}{\sin\dfrac{60^o}{2}} \\ =\dfrac{\sin 50^o}{\sin 30^o}=1.532$$\\$ Hence, the refractive index of the material of the prism is $1.532$$\\ Since the prism is placed in water, let \delta '_m be the new angle of minimum deviation for the same prism.\\ The refractive index of glass with respect to water is given by the relation:\\ \mu_g^w =\dfrac{\mu'}{\mu}\\ =\dfrac{\sin\dfrac{(A+\delta'_m)}{2}}{\sin \dfrac{A}{2}}\\ \sin\dfrac{(A+\delta'_m)}{2}=\dfrac{\mu'}{\mu}\sin \dfrac{A}{2}\\ \sin\dfrac{(A+\delta'_m)}{2}=\dfrac{1.532}{1.33}*\sin\dfrac{60^o}{2}\\ =0.5759\\ \dfrac{A+\delta'_m}{2}=\sin^{-1}0.5759=35.16^o\\ 60^o+\delta'_m=70.32^o\\ \therefore \delta'_m=70.32^o-60^o=10.32^o$$\\$ Hence, the new minimum angle of deviation is $10.32^o$

7   Double-convex lenses are to be manufactured from a glass of refractive index $1.55$ , with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 cm$ ?

##### Solution :

Refractive index of glass, $\mu= 1.55$$\\ Focal length of the double-convex lens, f = 20 cm$$\\$ Radius of curvature of one face of the lens $= R_ 1$$\\ Radius of curvature of the other face of the lens = R_ 2$$\\$ Radius of curvature of the double-convex lens $= R$$\\ \therefore R _1 = R and R_ 2 =- R$$\\$ The value of $R$ can be calculated as:$\\$ $\dfrac{1}{f}=(\mu-1)[\dfrac{1}{R_1}-\dfrac{1}{R_2}]\\ \dfrac{1}{20}=(1.55-1)[\dfrac{1}{R}+\dfrac{1}{R}]\\ \dfrac{1}{20}=0.55*\dfrac{2}{R}\\ \therefore R=0.55*2*20=22 cm$$\\ Hence, the radius of curvature of the double = convex lens is 22 cm$$\\$

8   A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam $12 cm$ from P. At what point does the beam converge if the lens is$\\$ (a) a convex lens of focal length $20 cm,$$\\ (b) a concave lens of focal length 16 cm? ##### Solution : In the given situation, the Object is virtual and the image formed is real. Object distance, u = +12 cm$$\\$ (a) Focal length of the convex lens,$f = 20 cm$$\\ Image distance = v$$\\$ According to the lens formula, we have the relation:$\\$ $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\ \dfrac{1}{v}-\dfrac{1}{12}=\dfrac{1}{20}\\ \dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{12}=\dfrac{3+5}{60}=\dfrac{8}{60}\\ \therefore v=\dfrac{60}{8}=7.5 cm$$\\ Hence, the image is formed 7.5cm away from the lens, toward its right.\\ (b) Focal length of the concave lens, f =- 16 cm$$\\$ Image distance $= v$$\\ According to the lens formula, we have the relation:\\ \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\ \dfrac{1}{v}=-\dfrac{1}{16}+\dfrac{1}{12}=\dfrac{-3+4}{48}=\dfrac{1}{48}\\ \therefore v=48 cm$$\\$ Hence, the image is formed $48 cm$ away from the lens, toward its right.

9   An object of size $3.0 cm$ is placed $14 cm$ in front of a concave lens of focal length $21cm.$ Describe the image produced by the lens. What happens if the object is moved further away from the lens?