Question

Consider the situation shown in figure. A plane mirror is fixed at a height h above the bottom of a beaker containing water (refractive index $\text{n}$) up to a height $\text{d}$. Find the position of the image of the bottom formed by the mirror.

• A. $\text{h - d}+\frac{\text{d}}{\text{n}}$
• B. $\text{h + d}+\frac{\text{d}}{\text{n}}$
• C. $\text{h - d}-\frac{\text{d}}{\text{n}}$
• D. $\text{d}+\frac{\text{d}}{\text{n}}$

Answer 1

The correct option is A $\text{h - d}+\frac{\text{d}}{\text{n}}$

$\text{n}=\frac{\text{Real depth}}{\text{Apparent depth}}$

When we look from above,

$\text{Apparent depth}=\frac{\text{Real depth}}{\text{n}}$

Change in depth of the beaker,

$\Delta \text{t}=\text{d}-\frac{\text{d}}{\text{n}}$

Distance of the bottom of the beaker from the plance mirror will be
$\text{h - Δt}=\text{h - d}+\frac{\text{d}}{\text{n}}$

Plane mirror forms the image at the same distance as that of object.
Therefore, the position of the image formed by the mirror will be

$\text{h - d}+\frac{\text{d}}{\text{n}}$