Question

A point source is placed at the bottom of a tank containing a transparent liquid (refractive index n) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is ______.

विकल्प

• `(pi H^2)/((n - 1))`

• `(pi H^2)/((n^2 - 1))`

• `(pi H^2)/(sqrt(n^2 - 1))`

• `(pi H^2)/((n^2 + 1))`

Answer 1

A point source is placed at the bottom of a tank containing a transparent liquid (refractive index n) to a depth H. The area of the surface of the liquid through which light from the source can emerge out is `bbunderline((pi H^2)/((n^2 - 1)))`.

Explanation:

Using basic trigonometry, the radius R of this circular patch is:

R = H tan θ_c

= `(H sin theta)/(cos theta_c)`

⇒ R = `(H sin theta_c)/(sqrt(1 - sin^2 theta_c))`

Since sin θ_c = `1/mu`

∴ R = `H/sqrt(mu^2 - 1)`

So, the area πR² = `(pi H^2)/(mu^2 - 1)`