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A glass slab of refractive index 5/3 and thickness 8 cm is placed over a point source of light on a flat surface. Light escapes from the top surface of the slab through a circular region of radius R cm. Find R.

1. 4 cm
2. 6 cm
3. 8 cm
4. 12 cm

Correct answer: 6 cm

Solution

Critical angle theta_c satisfies sin(theta_c) = 1/mu = 1/(5/3) = 3/5. So cos(theta_c) = 4/5, tan(theta_c) = 3/4. Light from the point source hits the top surface at various angles. The boundary of the illuminated circle corresponds to light hitting at exactly the critical angle. R = t * tan(theta_c) = 8 * (3/4) = 6 cm.

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