Two light beams fall on a transparent material block at point 1 and 2 with angle θ₁ and θ₂, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given : the distance between 1 and 2, d=4 √3 cm and θ₁=θ₂=cos ⁻¹(n₂/(2 n₁)). where refractive index of the block n₂> refractive index of the outside medium n₁, then the thickness of the block is ______ cm .

Solution

Physics sir · 4 steps

The angle is given from the surface, so convert it to the angle of incidence and Snell's law gives r = 30°.

1. Angle of incidence. In the figure, θ is measured from the surface, not from the normal. So the angle of incidence is i = 90° - θ, and sin i = cos θ = n₂/2n₁.
2. Apply Snell's law. Use n₁ sin i = n₂ sin r. Then n₁ × n₂/2n₁ = n₂ sin r, so sin r = 1/2 and r = 30° from the normal for both beams.
3. Use the symmetry. Both beams bend by the same angle towards each other, so they meet directly below the midpoint of 1 and 2. Each refracted ray covers a horizontal distance d/2 = 2√3 cm while going down the thickness t.

   tan r = (d/2)/t ⇒ t = 2√3/(tan 30°) = 2√3 × √3

4. So. The thickness of the block is t = 6 cm, so the answer is 6.

The common slip: 12. Taking the full separation d as the horizontal run of each ray, instead of d/2, gives 12 cm, not 6.

The idea to keep. Always check whether the angle is drawn from the surface or from the normal. Snell's law needs the angle from the normal, so a surface angle θ gives sin i = cos θ.

Answer from the NTA final answer key. Solution by PrepWiser.