State Snell’s law of refraction. A spherical surface of radius of curvature R separates a rarer and a denser medium as shown in the figure below : Complete the path of the ray of light, showing the formation of a real image. Derive the relation connecting object distance u , image distance v , radius of curvature R and the refractive indices μ 1 and μ 2 of the two media.

Snell's Law gives a relationship between the angle of incidence and the angle of refraction. According to Snell’s law of refraction of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, for the light passing through the given pair of media. This constant is equal to ratio of refractive index of the later medium to refractive index of the former medium. Let O be a point object in the rarer medium μ 1 . A ray of light OP incident normally on the refracting surface, another ray of light OA incident at a point A at an angle of incidence α , The two rays meet at a point I on the principal axis hence I is the real image of the point object. Using Snell's law at point A we get, μ 1 sin θ 1 = μ 2 sin θ 2 Since, θ 1 and θ 2 are so small that sin θ 1 = θ 1 and sin θ 2 = θ 2 . Here, θ 1 = α + v and v = β + θ 2 μ 1 θ 1 = μ 2 θ 2 ⇒ μ 1 α + v = μ 2 v - β v = μ 1 α + μ 2 β μ 2 - μ 1 Let, a r c P A = l Now, ∠ v = l R , ∠ α = l u and ∠ β = l v . Substituting this in above l μ 2 - μ 1 R = μ 1 l - u + μ 2 l v ⇒ μ 2 v - μ 1 u = μ 2 - μ 1 R Hence, the relation connecting object distance u , image distance v , radius of curvature R and the refractive indices μ 1 and μ 2 of the two media is μ 2 v - μ 1 u = μ 2 - μ 1 R .

Derive an expression for the magnifying power of an astronomical telescope in normal adjustment.

An astronomical telescope has the same construction as that of a compound microscope, but the focal length of the objective is larger when compared to the focal length of the eye piece. The light enters the objective lens from a distant object and at its second focal point a real image is formed in the tube. And the eyepiece magnifies this image producing a final inverted image. Here, β and α are so small that tan β = β and tan α = α . The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye. Thus, the magnifying power m = tan β tan α = β α ≈ h f e h f ∘ = f ∘ f e Hence, expression for the magnifying power of an astronomical telescope in normal adjustment is derived.

Show that the effective power of two thin lenses in contact is the sum of the power of each lens.

The power of a lens is specified as P = 1 f , where f is the focal length. Let two lenses L 1 and L 2 placed in contact. The focal length of the two lenses be f 1 and f 2 . u 1 = O C 1 , v 1 = I 1 C 1 u 2 = - I 1 C 1 , which is approximately equal to I 1 C 2 and v 1 = C 2 I which is approximately equal to C 1 I ∴ 1 u 1 + 1 v 1 = 1 f 1 and ∴ 1 u 2 + 1 v 2 = 1 f 2 1 O C 1 + 1 I 1 C 1 = 1 f 1 and 1 C 1 I - 1 I 1 C 1 = 1 f 2 1 O C 1 + 1 C 1 I = 1 f 1 + 1 f 2 = 1 F The focal length of the combined lens is given as 1 F = 1 f 1 + 1 f 2 F = f 1 f 2 f 1 + f 2 As the power of lens is given as P = 1 F = 1 f 1 + 1 f 2 . So, the power of combined lens is P n e t = P 1 + P 2 The power of any number of lenses in contact is equal to the algebraic sum of the power of two individual lenses.

With the help of a neat diagram, explain the working of a compound microscope. Obtain an expression for a magnifying power when the final image is formed at least distance of distinct vision.

Diagram of the compound microscope : Working: An A B object which is to be magnified is placed in front of the objective lens just beyond its principal focus so that its real, inverted, and magnified image A ' B ' is obtained beyond the centre of curvature ( C ) of the objective lens. The image A ' B ' becomes an object for the eyepiece. The position of image A ' B ' is adjusted such that it lies within the focal length of eyepiece. The eyepiece forms a virtual, erect, and magnified image A ' ' B ' ' of the object. Thus, the final image formed by the compound microscope is virtual, inverted, and magnified behind the object. Calculation of magnification: The ratio of the angle subtended by an image to the angle subtended by an object placed at least distance of distinct vision is called magnifying power of a microscope. If the angle subtended by an image is β and angle by an object placed at D is given by α . Therefore, magnifying power m = β α ⋯ ( i ) α = A B D Where, D is the least distance of distinct vision. As the eye is close to the eye lens, we can take the angle subtended at the lens for the human eye also. ∴ Angle made by an image P Q β = P Q E Q = A ' B ' E B ' or β = A ' B ' u e Where, u e = E B ' is the distance of the image from the eye lens. Substituting the value in the above equation m = A ' B ' u e A B D = A ' B ' A B × D u e ⋯ ( ii ) Now, in the similar triangle △ A ' O B ' & △ A O B A ' B ' A B = O B ' O B from the equation (ii) m = O B ' O B × D u e m = v 0 - u 0 × D u e (By sign convention) Where u 0 = O B is the distance of the object from an objective lens v 0 = O B ' is the distance between the objective lens to the image A ' B ' m = - v 0 D u 0 u e ⋯ ( iii ) When an image is formed at least distance of distinct vision : ∵ 1 f = 1 v - 1 u For eye lens 1 f e = 1 - v e - 1 - u e or 1 f e = 1 - v e + 1 u e but, v e = D 1 f e + 1 D = 1 u e D u e = D f e + 1 Substituting this value in the equation (iii) We get, m = - v 0 u 0 ( D f e + 1 ) This is the formula of magnification when the image is formed at least distance of distict vision.

Find an expression for combined focal length of two thin coaxial convex lenses placed in contact.

Let the focal length of the two lenses be f 1 and f 2 . u 1 = O C 1 , v 1 = I 1 C 1 u 2 = - I 1 C 1 , which is approximately equal to I 1 C 2 and v 1 = C 2 I which is approximately equal to C 1 I ∴ 1 u 1 + 1 v 1 = 1 f 1 (According to sign convention) and ∴ 1 u 2 + 1 v 2 = 1 f 2 (According to sign convention) 1 O C 1 + 1 I 1 C 1 = 1 f 1 and 1 C 1 I - 1 I 1 C 1 = 1 f 2 1 O C 1 + 1 C 1 I = 1 f 1 + 1 f 2 = 1 F The combined focal length of two thin lenses in contact is given by : F = f 1 f 2 f 1 + f 2

You are given two convex lenses of focal lengths 10 cm and 60 cm . To make a telescope, which of the two lenses will you use as object lens and which one as eye lens?

To make a telescope using the given convex lenses of focal length 10 cm and 60 cm , we will use the convex lens with focal length of 10 cm as eye piece, whereas the convex lens with focal length of 60 cm as objective lens.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Meghalaya-2020>Q 11

HARD

12th Meghalaya Board

IMPORTANT

Earn 100

Derive the relation between critical angle and refractive index.

Important Questions on Ray Optics and Optical Instruments

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Meghalaya-2020>Q 12

Why do prefer an objective of large aperture in a telescope?

MEDIUM

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Meghalaya-2020>Q 13

A small telescope has an objective lens of focal length

144 cm

and an eyepiece of focal length

6.0 cm

. What is the magnifying power of the telescope?

HARD

12th Meghalaya Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Meghalaya-2020>Q 14

Draw a ray diagram to show the refraction of light through a glass prism. Hence, obtain the relation for angle of deviation in terms of angle of incidence, angle of emergence and angle of prism. What is the refractive index of the material of the prism?

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