COMEDK 2023 Shift 1 Paper

For an equilateral prism where the angle of prism A is equal to 60∘, when the light undergoes minimum deviation, the angle of incidence (i) becomes equal to the angle of emergence (e) and both are equal to the angle of prism divided by 2 , i.e., i=e=‌

A

2

.
In this question, it is given that the angle of incidence is ‌

3

4

×A. When the light undergoes minimum deviation, for an equilateral prism, ‌

A

2

=‌

3

4

×A. Solving for A :
‌‌

A

2

=‌

3A

4

‌‌

1A

2

=‌

3A

4

This equation is incorrect, based on the information about minimum deviation. But for solving the refraction through a prism at minimum deviation and relating it to the velocity of light in the prism:
We use Snell's Law and the formula for refractive index. The refractive index n of the glass can be given by:
n=‌

sin‌( δ+A 2 )

sin‌(‌ A 2 )

At minimum deviation, i=e=‌

A

2

and the deviation δ is minimum. For an equilateral prism where A=60∘ :
n=‌

sin‌( δ+60∘ 2 )

sin‌(0∘.

sin‌(30∘)=0.5
Since the angle of incidence is ‌

3

4

A=‌

3

4

×60∘=45∘, and given that this is minimum deviation, let's skip forward with using the fact that it achieves symmetry at minimum deviation, we simplify calculations by using the angle condition:
From Snell's law at air-glass interface:
‌sin‌(i)=nsin‌(r)
‌sin‌(45∘)=nsin‌(30∘)
‌‌

√2

2

=n×0.5
‌n=‌

√2

1

=√2
The velocity of light in the glass prism, v, is related to the velocity in vacuum, c, by the refractive index:
v=‌

c

n

Given that n=√2 :
v=‌

c

√2

Thus, the answer is Option D:
‌

c

√2