Question

Find the area of the region to be covered on the face of glass cube of side 5 cm so that a spot at the centre cannot be seen from that face. μ = 1.5 for the glass cube.

Answer 1

we have to find the area of the region to be covered on the face of glass cube of side 5 cm so that a spot at the centre cannot be seen from that face. μ = 1.5 for the glass cube.

we know, radius of bright spot is given by, $\bf{R=\frac{h}{\sqrt{\mu^2-1}}}$

here h is edge length of cube and $\mu$ is refractive index of glass.

given, h = 5cm, $\mu$ = 1.5

so, R = 5cm/√(1.5² - 1) = 5/√(1.25) cm

now area of the region = πR²

= π {5/√(1.25)}²

= 25π/1.25

= 2500π/125

= 20π cm²

= 62.8 cm² [ as π = 3.14 ]

hence, area of the region to be covered on the face of glass cube of side 5cm so that a spot at the centre cannot be seen from that face = 62.8 cm²

Answer 2

we have to find the area of the region to be covered on the face of glass cube of side 5 cm so that a spot at the centre cannot be seen from that face. μ = 1.5 for the glass cube.

we know, radius of bright spot is given by, R=h/(√μ²-1)

here h is depth and μ is refractive index of glass.

given, h = 5/2 cm,  = 1.5

so, R = 5cm/(√(1.5² - 1)×2) = 5/(√(1.25)×2 cm)

now area of the region = πR²

= π {5/(√(1.25)×2)}²

= 25π/(1.25×4)

= 2500π/(125×4)

= 5π cm²

or

= 15.7 cm² [ as π = 3.14 ]

hence, area of the region to be covered on the face of glass cube of side 5cm so that a spot at the centre cannot be seen from that face = 15.7 cm²