The diagram below shows a fish in the tank and its image seen in the surface of water.

(a) Name the phenomenon responsible for the formation of this image.

(b) A double convex lens with refractive index μ₁ inside two liquids of refractive indices μ₂ and μ₃ are shown in the diagrams below. The refractive indices are such that μ₂ > μ₁ and μ₁ > μ₃

How would a parallel incident beam of light refract when it comes out of the lens in each of the cases shown above?

(1) in figure a.

(2) in figure b.

(a) Total internal reflection.

(b) According to the lens maker formula :

$\dfrac{1}{\text f} ∝ \left(\dfrac {\text {refractive index of 2nd medium}}{\text {refractive index of 1st medium} } - 1\right)$ ……(i)

(1) In figure a :

Refractive index of 1st medium = μ₃

Refractive index of 2nd medium = μ₁

Then ,from relation (i)

$\dfrac{1}{\text f} ∝ \left(\dfrac {\text μ$1 }{\text μ3} - 1\right)

As,

μ₁ > μ₃

$\Rightarrow \dfrac{\text μ$1}{\text μ3} \gt 1 \\[1 em] \Rightarrow \dfrac{1}{\text f} \gt 0 \\[1 em] \Rightarrow \text f \gt 0

Since, the resulting focal length is positive so this lens behaves as a converging lens.

Hence, a parallel incident beam of light converges when it comes out of the lens im figure a.

(2) In figure b :

Refractive index of 1st medium = μ₂

Refractive index of 2nd medium = μ₁

Then, from relation (i)

$\dfrac{1}{\text f} ∝ \left(\dfrac {\text μ$1 }{\text μ2} - 1\right)

As,

μ₁ < μ₂

$\Rightarrow \dfrac{\text μ$1}{\text μ2} \lt 1 \\[1 em] \Rightarrow \dfrac{1}{\text f} \lt 0 \\[1 em] \Rightarrow \text f \lt 0

Since, the resulting focal length is negative so this lens behaves as a diverging lens.

Hence, a parallel incident beam of light diverges when it comes out of the lens im figure b.