Question

A glass cube of edge 1 cm and refractive index 1.5 has a small spot at the centre. The area of cube face that must be covered to prevent the spot from being seen is

Answer 1

Answer:

Total area of the cube that is covered is given as 3.77 cm square

Explanation:

Let the angle of incidence at which light will show TIR is given as "theta"

so we will have

$n_1 sin\theta_1 = n_2 sin\theta$

so we will have

$1.5 sin\theta = 1 sin90$

$sin\theta = \frac{2}{3}$

so we will have

$\theta = 41.8$

now the radius of the circle from which light can come out of one face of the cube is given as

$tan\theta = \frac{r}{a/2}$

$tan41.8 = \frac{2r}{1 cm}$

$r = 0.45 cm$

total are of the cube to be covered is given as

$A = 6(\pi r^2)$

$A = 6(\pi \times 0.45^2)$

$A = 3.77 cm^2$

#Learn

Topic : Total internal reflection

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