IIT JEE , AIEEE Forum - Physics , Optics

jojojt

an equilateral prism deviates a ray through 23degree for two angles of incidence differing by 23degree.
find

of the prism.

ans sqrt43 / 5

joyfrancis

I'll tell you the method , tell me if you get the answer , it was pretty long so i left it but i'll tell you the logic.

Let the angle of incidence be

and angle of refraction in the first case be r1 and in the second case be r2.

Acc to the qn

= r₁+23 & ........(1)

+60 = r₂+23........(2)

By snells law

sin

/sinr₁=

........(3)

& sin(

+60)/sinr₂=

........(4)

Now see eqns 1,2,3 and 4 have 4 variables only which are r1,r2,

and

... it'll be long but i think you should get the answer by this.

jojojt

i think u went wrong somewhere joy.
can u please calculate and tell me the fina ans.

it is sqrt43 / 5

elessar_iitkgp

First understand this : Suppose a ray of light is incident on the prism at an angle i and emerges at an angle e. If it is made to be incident with an angle e, it will emerge at an angle i. In both cases, the angle of deviation would remain the same.

This is the concept to be applied here. So what the data of the problem is telling us is that, first it is incident at an angle i (and emergent at e) to give a deviation 23.
Next, its incident at i+23, and gives the same deviation.

Hence,
e = i+23

Now, i+e =

+A
i+i+23 = 23 +60
i=30
e=53

Suppose, when the angle incidence is i, the first refraction angle is r₁, and the second incidence angle is r₂.
Then,
sin i/sin r₁ =

and sin e/sin r₂ =

Equating these two,
sin r₁/sin r₂ = sin i/ sin e = (1/2) / (4/5) = 5/8
But r₁ + r₂ = A = 60
sin r₁/sin r₂ = sin (60 - r₂)/sin r₂ = 5/8
cot r₂ = 1/(4

3)
sin r₂ = (4

3)/7

sin r₁/sin r₂ = 5/8
sin r₁ = 5

3 / 14

As

= sin i/sin r₁ = (1/2) / 5

3 / 14 = 7/(5

3)
The answer isn't matching, but I'm pretty sure (unless I've made some calculation error) that the method is correct.

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