Question

Hi I'm Mainak.
a light ray is passes through a transparent sphere having radius R and refractive index 'mu'. The distance between the incident ray and a diameter of the sphere which is parallel to that ray is 'b'. What is the value of the deviation?

Do not post useless answers just by Googling.

Answer 1

Answer:

See ABTA SC-355

Explanation:

Answer 2

The value of the deviation is 2(sin⁻¹ b/R - sin⁻¹ b/Rμ)

Explanation:

The sine value of i is:

sin i = b/R

∴ i = sin⁻¹ b/R

Now, the refractive index is given by the formula:

μ = sin i/sin r

sin r = sin i/μ

On substituting the sine value of i, we get,

sin r = b/Rμ

∴ r = sin⁻¹ b/Rμ

Now, deviation is given by the formula:

δ = 2(i - r)

On substituting the values of 'i' and 'r', we get,

∴ δ = 2(sin⁻¹ b/R - sin⁻¹ b/Rμ)