Top 10 Allied and Healthcare Courses after Class 12 in India

Real Depth And Apparent Depth MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Real depth and Apparent depth is considered one the most difficult concept.

  • 14 Questions around this concept.

Solve by difficulty

The image of point $P$ when viewed from top of the slabs will be (both the top slab and bottom slab has a refractive index, $\mu=1.5$ )

A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface.  A small air bubble is 6 cm below the flat surface inside it along the axis.  The position of the image of the air bubble made by the mirror is seen :

 

An observer can see through a pinhole the top end of a thin rod of height h, placed as shown in the figure.  The beaker's height is 3h and its radius h.  When the beaker is filled with a liquid up to a height of 2h, he can see the lower end of the rod.  Then the refractive index of the liquid is 

An air bubble in a glass slab with a refractive index of 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is

An object is placed in front of a convex mirror at a distance of 50cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and plane mirror is 30cm, it is found that the images formed by two mirrors are along same line. Radius of curvature of mirror will be

A beam of light consisting of red, green and blue colour is incident on a right angled prism. The refractive indices of the material of the prism for the above red, green and blue wavelength are 1.39, 1.44 and 1.47 respectively. The prism will

The figure given below shows a beam of light converging at point P. When a concave lens of focal length 16 cm is introduced in the path of the beam at a place O shown by a dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value x will be equal to

.

JSS University Mysore 2025

NAAC A+ Accredited| Ranked #24 in University Category by NIRF | Applications open for multiple UG & PG Programs

A metal plate is lying at the bottom of a tank full of a transparent liquid. Height of tank is 100 cm but the plated appears to be at 45 cm above bottom. The refractive index of liquid is:

A fish in an aquarium approaches the left wall at a rate of $2.5 \mathrm{~m} / \mathrm{s}$ observes a fly approaching it at $8 \mathrm{~m} / \mathrm{s}$. If the refractive index of water is $(4 / 3)$, find the actual velocity of the fly.

NEET 2026 Free Mock Test with Solutions
Download the NEET 2026 Free Mock Test PDF with detailed solutions. Practice real exam-style questions, analyze your performance, and enhance your preparation.
Download EBook

A layer of oil 3 \mathrm{~cm} thick is floating on a layer of coloured water 5 \mathrm{~cm} thick. The refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36 / 7 \mathrm{~cm}. Find the refractive index of oil.

Concepts Covered - 1

Real depth and Apparent depth

Real depth and Apparent depth

Case 1.  When object is in denser medium and observer is in rarer medium.

If object and observer are situated in different medium then due to refraction, object appears to be displaced from it's real position.

Here O is the real position of the object and O' is the apparent position of the object as seen by the observer. 'h' is the real depth of the object from the surface of the water and h' is the

apparent depth of the object. $\mu_2$ is the density of the medium where the object is placed. $\mu_1$ is the density of the rarer medium.

$$
\frac{\mu_2}{\mu_1}=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{h}{h^{\prime}}
$$


Therefore Real depth > Apparent depth.

Apparent shift:

$$
d=h-h^{\prime}=\left(1-\frac{\mu_1}{\mu_2}\right) h
$$
 

Case 2.  Object is in rarer medium and observer is in denser medium.

$$
\frac{\mu_2}{\mu_1}=\frac{\text { Apparent depth }}{\text { Real depth }}=\frac{h^{\prime}}{h}
$$


Therefore apparent depth > real depth.
Apparent shift:

$$
d=\left(\frac{\mu_1}{\mu_2}-1\right) h
$$
 

 

Study it with Videos

Real depth and Apparent depth

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions