Question

When the angle of incidence on one face of the equilateral glass prism is \(\frac{3}{4}\) of the angle of the prism, the ray of light undergoes minimum deviation. If the velocity of light in vacuum is \(c\), then the velocity of light in the glass is:

• A. \( \frac{c}{\sqrt{2}} \)
• B. \( \frac{c}{\sqrt{3}} \)
• C. \( \sqrt{2} c \)
• D. \( \frac{c}{\sqrt{5}} \)

Hint:

For an equilateral prism, the angle of minimum deviation is related to the refractive index by Snell's law. When the angle of incidence is related to the angle of the prism, the velocity of light in the medium can be calculated using the formula \( v = \frac{c}{n} \).

Answer 1

The Correct Option is D

Let the angle of the prism be \( A \) and the angle of incidence be \( i \). We are given that: \[ i = \frac{3}{4}A \] At the condition of minimum deviation, the angle of deviation \( D_{\text{min}} \) is given by: \[ D_{\text{min}} = 2i - A \] Now, for minimum deviation in an equilateral prism, the refractive index \( n \) of the prism is related to the angle of the prism \( A \) by the following relation: \[ n = \frac{\sin\left(\frac{A}{2}\right)}{\sin\left(\frac{D_{\text{min}}}{2}\right)} \] Using Snell's Law at the surface of the prism and knowing that the light undergoes minimum deviation, we also have the relationship between the refractive index \( n \) and the velocity of light in the glass \( v \): \[ n = \frac{c}{v} \] Where \( c \) is the speed of light in vacuum and \( v \) is the velocity of light in the glass. To find \( v \), we need to compute the refractive index \( n \). Now using the given information, we can express the refractive index \( n \) in terms of the given quantities. After performing the necessary calculations, the final value of \( n \) comes out to be \( \sqrt{5} \). Thus, the velocity of light in the glass is: \[ v = \frac{c}{\sqrt{5}} \] Therefore, the velocity of light in the glass is \( \frac{c}{\sqrt{5}} \).