Question

When a ray is refracted through a prism, then
(a) ∠i=∠δ (b) ∠ i=∠e+∠δ
(c) ∠δ= ∠e (d) ∠i > ∠r

Answer 1

Answer:

Correct option is

C

∠A+∠D=∠i+∠e

D=ir1+e−r2 .... (1)

D=i+e(r1+r2)

Again, in quadrilateral ALOM,

∠ALO+∠AMO=2rt∠s [Since, ∠ALO=∠AMO=90]

So, ∠LAM+∠LOM=2rt∠s [Since, Sum of four ∠s of a quadrilateral = 4 rt∠s] .... (2)

Also in △LOM,

∠r1+∠r2+∠LOM=2rt∠s .... (3)

Comparing (2) and (3), we get

∠LAM=∠r1+∠r2

A=∠r1+∠r2

Using this value of ∠A, equation (1) becomes,

D=i+e−A

or ∠i+∠r=∠A+∠D

solution

Explanation:

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Answer 2

A ray is refracted through a prism, then

∠A+∠D=∠i+∠e

D=ir1+e−r2 .... (1)

D=i+e(r1+r2)

Again, in quadrilateral ALOM,

∠ALO+∠AMO=2rt∠s [Since, ∠ALO=∠AMO=90]

So, ∠LAM+∠LOM=2rt∠s [Since, Sum of four ∠s of a quadrilateral = 4 rt∠s] .... (2)

Also in △LOM,

∠r1+∠r2+∠LOM=2rt∠s .... (3)

Comparing (2) and (3), we get

∠LAM=∠r1+∠r2

A=∠r1+∠r2

Using this value of ∠A, equation (1) becomes,

D=i+e−A

or ∠i+∠r=∠A+∠D

Hence, we get the correct option is  C.