Question

The figures above show the refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence
in water is 45° with the normal to a water-glass interface.​

Answer 1

Answer:

${sin}^ - 1} \frac{sin35}{sin47 \sqrt{2} }$

Explanation:

Using sine law . In these type of 3 medium,given refractive index of 2 at a time questions . Always write sine law eq for each combination at a time. To find required divide or multiply two equations(constant) and equate it with variable to be find expression formed from sin law. Take inverse of T value of required angle.

Answer 2

Given :

✴ First case :

▪ Angle of incident = 60°

▪ Angle of refraction = 35°

✴ Second case :

▪ Angle of incident = 60°

▪ Angle of refraction = 47°

✴ Third case :

▪ Angle of incident = 45°

To Find :

▪ Angle of refraction in third case.

SoluTion :

We know that,

Refractive index of air = 1

✴ Applying snell's law in first case:

→ n¹sinΦ¹ = n²sinΦ²

where,

n¹ denotes refractive index of air

n² denotes refractive index of glass

Φ¹ denotes angle of incident

Φ² denotes angle of refraction

→ (1)(sin60°) = n²(sin35°)

→ √3/2 = n² × 1/√3

→ n² = (√3 × √3)/2

→ n² = 3/2

✴ Applying snell's law in second case:

→ n¹sinΦ¹ = n³sinΦ³

where

n³ denotes refractive index of water

→ (1)(sin60°) = n³(sin47°)

→ √3/2 = n³ × 3√3/8

→ n³ = 4/3

✴ Applying snell's law in third case:

→ n³sinΦ³ = n²sinΦ²

→ (4/3)(sin45°) = (3/2)(sinΦ²)

→ (4/3)(1/√2) = (3/2)(sinΦ²)

→ sinΦ² = 4√2/9

→ Φ² = 38.79°