Question

A ray is incident at an angle of incidence i on one surface of
a prism of small angle A and emerges normally from the
opposite surface. If the refractive index of the material of
prism is m, the angle of incidence i is nearly equal to
(a)A/μ
(b) A/2μ
(c) μ A (d)μA/2

Answer 1

Answer:Given data:

The angle of incidence =i

The angle of the prism =A

The angle of emergence =e

the refractive index of the material of the prism is μ.

For a small angle prism, angle of deviation, δ=(μ−1)A .......(i)

Since, the ray emerges normally, ∴e=0

By the relation,

A+δ=i+e

We have, i=A+δ

or δ=(i−A) .........(ii)

From Eqs. (i) and (ii)

(i−A)=(μ−1)A

⇒i=μA

Explanation:please Mark me as brainlist I really need it

Answer 2

Answer:    α=180  

Explanation:

0

−90  

0

−A=90−A

   r=90−α=A

from snell's law

     μ  

1

​  

sini=μ  

2

​  

sinr

     1sini=μsinA

for small angle

     

i=μA

​