For a thin lens, prove that 1 f = μ - 1 1 R 1 - 1 R 2

The given equation is of the Lens Maker's Formula, to prove this, let us take any lens made up of two surfaces A B C and A D C of radii of curvatures R 1 and R 2 . A point object is placed on the principal axis at O . Image of an object O is formed at I due to A B and this image acts as an object for the second surface A D and its image produced at I 1 . Let the distance of the object O from the lens is u and the image I 1 is at distance v 1 , and Image I is at distance v . From the first surface A B C image of O is obtained at I , that way by the formula for refraction at spherical surface, μ - 1 R 1 = μ v 1 - 1 u . . . . . . . . . . . . i Here, μ is the refractive index which is the ratio of the speed of light in a vacuum to its speed in a specific medium. From the second surface A D C image of I is obtained at I 1 ,by the formula for refraction at spherical surface, 1 μ - 1 R 2 = 1 μ - 1 v - 1 v 1 . . . . . . . . . . i i Here, μ = 1 u , u = v 1 for A D surface because light ray passing from denser to rarer medium Now, on multiplying μ with equation i i 1 v - μ v 1 = - 1 - μ R 2 . . . . . . . . . . . . i i i Now, adding equation i & i i i 1 v - 1 u = μ - 1 R 1 + 1 - μ R 2 . . . . . . . . . . i v 1 v - 1 u = μ - 1 R 1 - μ - 1 R 2 We know that when u = ∞ then v = f 1 f - 1 ∞ = μ - 1 1 R 1 - 1 R 2 Now, we know that 1 ∞ = 0 So, 1 f = μ - 1 1 R 1 - 1 R 2 Hence, Proved.

Draw the ray diagram to show the formation of the image by compound microscope and obtain an expression for the magnifying power when the final image is formed at least distance of distinct vision.

Compound microscope consists of two convex lenses separated by a distance. The lens near the object is called objective and the lens near the eye is called eye piece. The objective lens has small focal length and eye piece has of larger focal length. The object placed near the objective lens produces an image and this image acts as object for the eye piece lens. The position of the eye piece lens is adjusted such that it produces virtual and magnified image of the object. The magnifying power of a microscope is the ratio of the angle subtended by image to the angle subtended by image placed at least distance of distinct vision. Therefore, from the above diagram, m = β α ….. (1) ∴ α = A B D …… (2) Where D is the least distance of distinct vision i.e. the distance between point O and Point B Now, the angle made by image A ' B ' ' can be given by, β = A ' B ' E B ' Let, u e be the distance between image and eyes piece Thus, β = A ' B ' u e ……. (3) Now, from (1), (2) and (3) We get, m = A ' B ' u e × D A B From the diagram above we can say that, A ' B ' A B = O B ' O B Substitute this in the above equation, m = O B ' O B × D u e Here, O B and O B ’ is the distance of object and image from point O . Therefore, m = v 0 u 0 × − D − u e ……. (4) On solving, m = v 0 D u 0 u e Using the lens formula, we know, 1 f = 1 v - 1 u The lens formula for the microscope: 1 f e = 1 v e - 1 u e Since the distance D is positive. Therefore, 1 - f e = 1 D - 1 u e Rearranging the terms, We get, 1 u e = 1 f e + 1 D On solving, D u e = D f e + 1 ….. (5) Now, from (4) and (5), We get, m = v 0 u 0 1 + D f e Therefore, the magnification power of a compound microscope is given by, m = v 0 u 0 1 + D f e

For a thin lens, prove that 1 f = μ - 1 1 R 1 - 1 R 2

The given equation is of the Lens Maker's Formula, to prove this, let us take any lens made up of two surfaces A B C and A D C of radii of curvatures R 1 and R 2 . A point object is placed on the principal axis at O . Image of an object O is formed at I due to A B and this image acts as an object for the second surface A D and its image produced at I 1 . Let the distance of the object O from the lens is u and the image I 1 is at distance v 1 , and Image I is at distance v . From the first surface A B C image of O is obtained at I , that way by the formula for refraction at spherical surface, μ - 1 R 1 = μ v 1 - 1 u . . . . . . . . . . . . i Here, μ is the refractive index which is the ratio of the speed of light in a vacuum to its speed in a specific medium. From the second surface A D C image of I is obtained at I 1 ,by the formula for refraction at spherical surface, 1 μ - 1 R 2 = 1 μ - 1 v - 1 v 1 . . . . . . . . . . i i Here, μ = 1 u , u = v 1 for A D surface because light ray passing from denser to rarer medium Now, on multiplying μ with equation i i 1 v - μ v 1 = - 1 - μ R 2 . . . . . . . . . . . . i i i Now, adding equation i & i i i 1 v - 1 u = μ - 1 R 1 + 1 - μ R 2 . . . . . . . . . . i v 1 v - 1 u = μ - 1 R 1 - μ - 1 R 2 We know that when u = ∞ then v = f 1 f - 1 ∞ = μ - 1 1 R 1 - 1 R 2 Now, we know that 1 ∞ = 0 So, 1 f = μ - 1 1 R 1 - 1 R 2 Hence, Proved.

HARD

12th Chhattisgarh Board

IMPORTANT

Earn 100

Describe the terrestrial telescope under the following points: Derivation of formula for magnifying power when the image is formed at least distance of distinct vision.

Important Questions on Ray Optics and Optical Instruments

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2018>Q 7

Describe the terrestrial telescope under the following points: Ray diagram of the formation of image.

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 8

A candle is placed at

12 cm

distance from a concave mirror. Radius of curvature of the mirror is

10 cm

. Find the position, nature and magnification of the image.

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 9

Draw a labelled diagram depicting the formation of image by an Astronomical Telescope in normal adjustment (when final image is formed at infinity

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 10

With the help of a ray diagram, show the formation of the image of a point object by the refraction of light at a spherical convex surface separating two media of refraction indices

μ_{1}

and

μ_{2}

(

μ_{2} > μ_{1}

) respectively, Using this diagram derive the relation:

\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}

Where used symbols have their usual meanings

HARD

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>chhattisgarh board-2019>Q 11

Draw the ray diagram showing the refraction of monochromatic light through a glass prism of refraction angle

A

. Hence, derive the formula for the refractive index

μ

of glass prism in terms of angle of prism and angle of minimum deviation. Also, draw

i - δ

curve of prism.

EASY

12th Chhattisgarh Board

IMPORTANT

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

Which mirror has an infinite focal length?

EASY

12th Chhattisgarh Board

IMPORTANT

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

What is optical fibre? Write its two important uses.

HARD

12th Chhattisgarh Board

IMPORTANT

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

For a thin lens, prove that

\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

Describe the terrestrial telescope under the following points: Derivation of formula for magnifying power when the image is formed at least distance of distinct vision.

Important Questions on Ray Optics and Optical Instruments

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