A large glass slab (μ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. The light from the source emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?
Physics · Mixed Chapter
JEE Mains

Question

A large glass slab (μ = 5/3) of thickness 8 cm is placed over a point source of light on a plane surface. The light from the source emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Rays can emerge from glass into air only up to the critical angle θc, with sinθc=1/μ=3/5. Step 2: Hence cosθc=4/5 and tanθc=3/4. Step 3: With slab thickness t=8 cm, the radius of the emergence circle is R=t tanθc=8×3/4=6 cm. Answer: 6.