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Real Depth And Apparent Depth - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

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

  • 16 Questions around this concept.

Solve by difficulty

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 :

 

A microscope is focused on an object at the bottom of a bucket. If liquid with a refractive index \frac{5}{3} is poured inside the bucket, then the microscope has to be raised by 30 \mathrm{~cm} to focus the object again. The height of the liquid in the bucket is :

A bulb is placed at a depth of \mathrm{2\sqrt{7}\ m} in water \mathrm{\left ( \mu _{w}=4/3 \right )} and a floating opaque disc is placed over the bulb so that the bulb is not visible from the surface. What is the minimum diameter of the disc?

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

In a lake, a fish rising vertically to the surface of water uniformly at the rate of \mathrm{3\ m/s}, observes a bird diving vertically towards the water at the rate of \mathrm{9\ m/s}. The actual velocity of the dive of the bird is (given, refractive index of water \mathrm{=4/3})

The apparent depth of water in the cylindrical water tank of diameter \mathrm{2R} cm is reducing at the rate of \mathrm{x} \mathrm{cm/min} when water is being drained out at a constant rate. The amount of water drained in \mathrm{cc/min} is ( \mathrm{n_{1}=} refractive index of air, \mathrm{n_{2}=} refractive index of water)

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'}{h}.   

Therefore apparent depth > real depth.

Apparent shift:

d=\left(\frac{\mu _1}{\mu_2}-1\right) h

 

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Real depth and Apparent depth

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