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(cm^2)

Answer 1

Answer:

Area of one face of the cube that is to be covered is 0.63 cm^2

Explanation:

As we know that light will not move out of the face of the cube when it will suffer total internal reflection

So we will have

$\mu_1 sin\theta_1 = \mu_2 sin\theta_2$

here we have

$1.5 sin\theta = 1 sin90$

$\theta = 41.8 degree$

now let say the radius of the part of the cube from which light is coming out is given as "r"

then we will have

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

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

$r = 0.5 tan41.8$

$r = 0.45 cm$

so the area of the face of the of the cube is given on one side that should be covered

$A = \pi r^2$

$A = \pi (0.45)^2$

$A = 0.63 cm^2$

#Learn

Topic : Total internal reflection

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