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 = c…

Question

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)/(2n₁)), where refractive index of the block n₂ > refractive index of the outside medium n₁, then the thickness of the block is cm.

Accepted Answer

6. θ is measured from the surface, so the angle of incidence is i = 90°-θ and sin i = cos θ = (n₂)/(2n₁). Snell's law: n₁ sin i = n₂ sin r gives sin r = 1/2, r = 30°. By symmetry point 3 is below the mid-point of 1 and 2, so each refracted ray covers d/2 = 2√3 cm sideways. Thickness t = (d/2)/(tan 30°) = 2√3×√3 = 6 cm. → 6

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