1) A light ray propagates in a material with index of refraction 1.13, strikes a

Question

1) A light ray propagates in a material with index of refraction 1.13, strikes an interface, and passes into a material whose index of refraction is 1.31. The angle of incidence at the interface is 22.5°. Find the angle of refraction.

2) a) You are given a spherical mirror and wish to determine its properties. You place an object on its axis, 36.3 cm in front of it, and discover that the mirror creates a virtual image located 10.7 cm from the mirror. Find the mirror's focal length.

b) Calculate the mirror's radius of curvature.

Explanation / Answer

(1) given

u1 = 1.13

u2 = 1.31

Q1 = 22.5

we know that the snels law

u1sinQ1 =u2sinQ2

sinQ2 =u1sinQ1/u2

sinQ2 =1.13xsin22.5/1.31=-0.420234

Q2 = -0.4337 Ans

(2)

(a) given that

U = -36.3 cm

V = 10.7 cm because the image is virtual

1/f=1/v+1/u

1/f=1/10.7-1/36.3

1/f = 100/10.7-100/36.3

1/f = 9.34579 - 2.75482

1/f =6.590974

f=1/6.590974 =0.15172 m=15.17cm ans

(b)

we know that the focal length depend on the radius of curvature as follows

f=r/2

r =2f

r=2x15.172= 30.344cm ans

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