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Refraction Of Light Through Glass Slab MCQ - Practice Questions with Answers

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

Quick Facts

  • Refraction Through A Glass Slab is considered one the most difficult concept.

  • 12 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 convex lens is put 10cm from a light source and it makes sharp images on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is:

An air bubble in a glass slab with refractive index 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

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The mirror of length L moves horizontally as shown in the figure with a velocity v. The mirror is illuminated by a point source of light 'P' placed on the ground. The rate at which the length of the light spot on the ground increases is:

Two beams of light are incident normally on water (R.I. = 4/3). If the beam 1 passes through a glass \mathrm{(R.I. =3 / 2)} slab of height h as shown in the figure, the time difference for both the beams for reaching the bottom is

A convex lens focuses a distant object on a screen placed 10 cm away from it. A glass plate \mathrm{(n=1.5)} of thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen?

A rectangular glass slab ABCD, of refractive index \mathrm{n}_{1}, is immersed in water of refractive index \mathrm{n_{2}\left(n_{1}>\right.\mathrm{n}_{2} )}.A ray of light is incident on the surface AB of the slab as shown. The maximum value of the angle of incidence \mathrm{\alpha_{\max }}, such that the ray emerges only from the surface CD is given by

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A diverging beam of light from a point source S having divergence angle \alpha, falls symmetrically on a slab as shown. The angle of incidence of the two extreme rays are equal.
If the thickness of the glass slab is t and the refractive index n then the divergence angle of the emergent beam is

 

A ray of light is incident on the left vertical face of a glass cube of refractive index \mu_{2},as shown in figure. The plane of incident is the plane of the page, and the cube is surrounded by liquid \left(\mu_{1}\right).What is the largest angle of incidence \theta_{1} for which total internal reflection occurs at the top surface? 

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Concepts Covered - 2

Refraction Through A Glass Slab

 

Consider an object O placed a distance d in front of a glass slab of thickness "t" and refractive index  \mu. The observer is on
the other side of the slab. A ray of light from the object first refracts at the surface (1) then refracts at the surface ( 2 ) before reaching the observer as shown in the above figure.  

So for the refraction at the surface (1)

\text { Apparent depth, } d_1'=\frac{d_{\text {real }}}{n_{\text {relative }}}=\frac{d}{\left(\frac{n_{\text {incident }}}{n_{\text {reftaction }}}\right)}= \frac{d}{\frac{1}{\mu }}=d\mu

 

Similarly for the refraction at the surface (2)

\text { Apparent depth, } d_2'=\frac{d_{\text {real }}}{n_{\text {relative }}}=\frac{d_1'+t}{\left(\frac{n_{\text {incident }}}{n_{\text {reftaction }}}\right)}= \frac{d_1'+t}{\frac{\mu }{1 }}= \frac{d_1\mu +t}{\mu }

 

 As you observe, The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges out parallel to its incident direction.

i.e. the ray undergoes no deviation ( \delta =0).

the object appears to be shifted towards the slab by the distance known as apparent shift or Normal shift.

And the apparent shift= OA-I2A

I.e \begin{array}{l}{\text { Apparant shift }=t\left\{1-\frac{1}{\mu}\right\}} \\ \end{array}

If the slab is placed in the medium of refractive index \mu _{sur}

then \begin{array}{l}{\text { Apparant shift }=t\left\{1-\frac{\mu _{sur}}{\mu}\right\}} \\ \end{array}

Lateral Displacement Of Emergent Ray Through A Glass Slab

In the above figure Incident, ray AO is an incident on the EF surface of the slab at an angle of incident i, and PB is the emergent ray emerging out of the HG surface of the slab.

for the surface EF

Applying Snell's law at  the surface EF and HG  

\begin{array}{l}{\mu_{a} \sin i=\mu \sin r \quad \text { and } \quad \mu \sin r^{\prime}=\mu_{a} \sin e} \\ Using \ \ {r^{\prime}=r \quad \text { and } \quad \mu_{a}=1, \text { we get }} \\ {\sin i=\sin e \text { or } e=i}\end{array}

i.e the emergent ray is parallel to the incident ray.

If  PQ is the perpendicular dropped from P on the incident ray produced.

Then PQ=d is known as lateral displacement which is given as

\begin{aligned} d &=P Q=O P \sin (i-r)=\frac{O M}{\cos r} \sin (i-r) =\frac{t \sin (i-r)}{\cos r} \end{aligned}

  • \text { If } i \text { is very small, } r \text { is also very small, then } \ \ d=\left(1-\frac{1}{\mu}\right) t i 

 

 

Study it with Videos

Refraction Through A Glass Slab
Lateral Displacement Of Emergent Ray Through A Glass Slab

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