Derive an equation δ = i + e - A for an equilateral prism and using it, obtain an equation for the refractive index ( n ) of the material of the prism.

Let light ray PQ be the incident ray. The angles of incidence and refraction at the first face AB are i 1 and r 1 respectively. QR is the path followed by the light inside the prism. The angle of incidence and refraction at the second face AC are r 2 and i 2 respectively, RS is the emerging ray from the second face. Angle i 2 is also called the angle of emergence ( e ) . δ is the angle of deviation formed between the direction of the incident ray PQ and the emergent ray RS . The two normals are drawn at the point of incidence Q and emergence R are QN and RN . They meet at the point N . The incident ray and emergent Ray meet at a point M . The angle of deviation δ 1 at the surface AB is, ∠ RQM = δ 1 = i 1 - r 1 . The angle of deviation δ 2 at the surface AC is, ∠ QRM = δ 2 = i 2 - r 2 . The total angle of deviation δ = δ 1 + δ 2 . By substituting the values of δ 1 and δ 2 , δ = i 1 - r 1 + i 2 - r 2 . In the quadrilateral AQNR two of the angles at the vertices Q and R are right angles therefore, the sum of the other angle of the quadrilateral is 180 ° . Hence, ∠ A + ∠ QNR = 180 ° - - - - - - - - - - ( i ) . Now, from the triangle QNR , r 1 + r 2 + ∠ QNR = 180 ° - - - - - - - - - - ( i i ) Comparing equation ( i ) and ( i i ) r 1 + r 2 = A Now, by substituting the obtained value in δ = i 1 - r 1 + i 2 - r 2 . δ = i 1 + i 2 - A . This is the angle of deviation for an equilateral prism. As, i 2 = e (angle of emergence) Therefore, it can be rewritten as δ = i 1 + e - A . Now, let us obtain the refractive index of the material of the prism. At minimum deviation, i 1 = i 2 = i and r 1 = r 2 = r . Now using the equation for the angle of deviation i.e., δ = i 1 + i 2 - A = 2 i - A or i = A + δ 2 . And, as r 1 + r 2 = A = 2 r = A or r = A 2 . Now according to Snell's law n = sin i sin r . Now substituting the value of i and r in above expression. n = sin A + δ 2 sin A 2 . Therefore, the refractive index of the material of the prism can be obtained by using this expression i.e., n = sin A + δ 2 sin A 2

EASY

12th Gujarat Board

IMPORTANT

Earn 100

Obtain relation between focal length and radius of curvature for concave mirror.

Important Questions on Ray Optics and Optical Instruments

HARD

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Gujarat Board-2018>Q 6

A converging lens of focal length $15 cm$ and converging mirror of focal length $20 cm$ and place with principal axis coinciding . A point object is placed at a distance $12 cm$ from the lens. Refracted ray from the lens gets reflected from the mirror and again refracted by the lens is parallel to the principal axis. Find the distance between the mirror and the lens.

EASY

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Gujarat Board-2019>Q 7

Detailed information can be obtained by the OIL Immersion objective of a microscope because the objective has _______.

EASY

12th Gujarat Board

IMPORTANT

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

The ratio of resolving power of telescope, when lights of wavelengths

5000 \overset{\circ}{A}

and

4000 \overset{\circ}{A}

are used, is _____

EASY

12th Gujarat Board

IMPORTANT

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

For plane mirror, the value of magnification

m = 1

, then its focal length

f =

_____.

EASY

12th Gujarat Board

IMPORTANT

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

An object is placed at a distance of

25 cm

on the axis of a concave mirror having focal length

20 cm

. What will be the lateral magnification of an image?

MEDIUM

12th Gujarat Board

IMPORTANT

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

Depth of a completely filled well is

11 m

and refractive index of water is

1.33

. If viewed normally from the top, by how much height would the bottom of the well appears to be shifted up?

EASY

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Gujarat Board-2019>Q 12

Which of the following is responsible for the glittering of a diamond?

EASY

12th Gujarat Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR PHYSICS>Chapter 9 - Ray Optics and Optical Instruments>Gujarat Board-2019>Q 13

A defect of vision in which lines in one plane of an object appears in focus while those in another plane are out of focus is called ______.

Obtain relation between focal length and radius of curvature for concave mirror.

Important Questions on Ray Optics and Optical Instruments

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