Refractive index of water and glasses are 4/5 and 5/3.A light ray is going to water from glass.Then its critical angle will b?

To find the critical angle when light travels from glass to water, we can use Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media. The critical angle is the angle of incidence in the denser medium (glass, in this case) that results in an angle of refraction of 90 degrees in the less dense medium (water).

Understanding the Refractive Indices

The refractive index (n) is a measure of how much light bends when it enters a material. For this problem, we have:

• Refractive index of glass (n_glass) = 5/3
• Refractive index of water (n_water) = 4/5

Applying Snell's Law

Snell's Law states that:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

Where:

• n₁ = refractive index of the first medium (glass)
• θ₁ = angle of incidence in the first medium
• n₂ = refractive index of the second medium (water)
• θ₂ = angle of refraction in the second medium

Finding the Critical Angle

At the critical angle (θ_c), the angle of refraction (θ₂) is 90 degrees. Thus, sin(θ₂) = sin(90°) = 1. We can rewrite Snell's Law for this scenario:

n_glass * sin(θ_c) = n_water * 1

Substituting the values of the refractive indices:

(5/3) * sin(θ_c) = (4/5)

Solving for the Critical Angle

Now, we can isolate sin(θ_c):

sin(θ_c) = (4/5) / (5/3)

This simplifies to:

sin(θ_c) = (4/5) * (3/5) = 12/25

Calculating the Critical Angle

To find θ_c, we take the inverse sine (arcsin) of 12/25:

θ_c = arcsin(12/25)

Using a calculator, you would find:

θ_c ≈ 28.4 degrees

Therefore, the critical angle for light transitioning from glass to water is approximately 28.4 degrees. This means that if the angle of incidence in the glass exceeds this critical angle, total internal reflection will occur, and the light will not pass into the water.