Home
Class 12
PHYSICS
In an experiment for determination of re...

In an experiment for determination of refractive index of glass of a prism by `i-delta`, plot it was found thata ray incident at angle `35^circ`, suffers a deviation of `40^circ` and that it emerges at angle `79^circ`. In that case which of the following is closest to the maximum possible value of the refractive index?

A

`1.6`

B

`1.7`

C

`1.8`

D

`1.5`

Text Solution

Verified by Experts

The correct Answer is:
D
`i = 35^(@) , e = 79^(@) and delta = 40^(@)`
`delta = i + e - A `
`rArr 40^(@) = 35^(@) + 79^(@) - A rArr A = 74^(@)`
Since, `i ne e ` hence ` delta_(m)` will be less than `40^(@)`
`(sin ((delta_(m)+A)/(2)))/(sin A//2) = (sin ((40 + 74)/(2)))/(sin (74//2)) = (sin 57)/(sin 37) =1.4`
Closest answer is 1.5 from options
Promotional Banner

Similar Questions

Explore conceptually related problems

A prism of refractive index sqrt(2) has a refracting angle of 60^(@) . At what angle must a ray incident on it so that it suffers minimum deviation ?

The angle of a prism is 30^(@) . The rays incident at 60^(@) on one refracting face suffer a deviation of 30^(@) . Then the angle of emergence is :

A light ray is incident perpendicularly to one face of a 90^circ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45^circ , we conclude that the refractive index n

A ray incident a 15^(@) on one refracting surface of a prism of angle 60^(@) , suffers a deviation of 55^(@) . What is the angle of emergence

Light rays from a source are incident on a glass prism of index of refraction mu and angle of prism alpha At near normal incidence, the angle of deviation of the emerging rays is

What should be the minimum value of refractive index of a prism, refracting angle A, so that there is no emergent ray irrespective of the angle of incidence?