Question

There is an air bubble inside water of refractive index $4/3$ . At what distance from the air bubble should a point object so as to form a real image at the same distance from the bubble?
(A) $R$
(B) $2R$
(C) $4R$
(D) Not possible

Answer 1

Hint:The trapping of the air in the liquid layers forms the bubble and this acts as the diverging lens due to its bulged spherical surface. The real image is possible only when all the light from the object meets at one point after the reflection. Inside the water, the bubble acts as the concave lens and outside the water, it acts as a convex lens.

Complete step by step solution:
It is given that the
Refractive index of the bubble, $n = \dfrac{4}{3}$
It is known that the bubbles are formed when some amount of air is enclosed or entrapped in the water droplets. This bubble is formed due to the force of surface tension all over the surface by acting in the centre of the bubble and this force is resisted by the pressure of the air towards the air surface. The air is of less density and the water surface is of high density, when these two joins it forms the bulging spherical surface. This bulging surface acts as the diverging lens. The diverging lens forms the virtual and the erect image, and it does not form the real image.

Thus the option (D) is correct.

Note:The difference between the virtual and the real image are as follows.
Virtual image seen from the mirror and the real image from the convex and the concave lens.
Virtual image formed only when the light rays from the object meet at a point backwards after the reflection from a mirror. Whereas the real image is the light rays intersect at a point after reflection.