Questions on reflection and refraction at plane surfaces are interesting to answer, but those involving spherical surfaces may pose some problems to you. Let us discuss some questions involving reflection and refraction at plane surfaces. (We will discuss questions involving spherical surfaces later).

The following MCQ comes under normal refraction:

Here 4 cm and 2 cm are the apparent distances of the air bubble on looking through the two faces. If ‘d’ is the real distance, we can write the refractive index (n) as n = d/4 = (9-d)/2. [We have written n = (Real distance)/(Apparent distance) for the two cases]. From this, d = 6 cm so that n = 6/4 = 1.5.

Suppose there are many layers of thickness t

The following MCQ comes under normal refraction:

**A transparent cube of edge 9cm contains a small air bubble which appears to be at a distance of 4 cm when viewed normally through one face and at a distance of****2 cm when viewed normally through the opposite face.****The refractive index of the material of the cube is**

(a) 1.4 (b) 1.45 (c) 1.5 (d) 1.55 (e) 1.6(a) 1.4 (b) 1.45 (c) 1.5 (d) 1.55 (e) 1.6

Here 4 cm and 2 cm are the apparent distances of the air bubble on looking through the two faces. If ‘d’ is the real distance, we can write the refractive index (n) as n = d/4 = (9-d)/2. [We have written n = (Real distance)/(Apparent distance) for the two cases]. From this, d = 6 cm so that n = 6/4 = 1.5.

Suppose there are many layers of thickness t

_{1}, t_{2}, t_{3}etc of different immiscible liquids of refractive indices n_{1}, n_{2}, n_{3}etc in a vessel. What will be the apparent depth? You can work it out using Snell’s law. The total apparent depth turns out to be equal to the sum of the individual apparent depths:**Total apparent depth = (t**

Now consider the following MCQ:_{1}/n_{1}) + (t_{2}/n_{2)}+ (t_{3}/n_{3}) +…etc.**A jar is half filled with a liquid of refractive index μ and the other half is filled with another immiscible liquid of refractive index 1.2μ. The apparent depth of the jar is then two thirds the actual depth. Then, the refractive index of the rarer liquid is**

(a) 1.6 (b) 1.575 (c) 1.55 (d) 1.475 (e) 1.375(a) 1.6 (b) 1.575 (c) 1.55 (d) 1.475 (e) 1.375

If the thickness of the liquid layers are ‘t’ each, the actual depth of the jar is 2t. The apparent depth as given in the question is (2/3)×2t =4t/3 so that we can write,

4t/3 = t/μ + t/1.2μ from which μ = 1.375.

Here is a simple question which appeared in Kerala Medical Entrance 2005 test paper:

4t/3 = t/μ + t/1.2μ from which μ = 1.375.

Here is a simple question which appeared in Kerala Medical Entrance 2005 test paper:

**A glass slab of thickness 3 cm and refractive index 3/2 is placed on ink mark on a piece of paper. For a person looking at the mark from a distance 5 cm above it, the distance of the mark will appear to be**

(a) 3 cm (b) 4 cm (c) 4.5 cm (d) 5 cm (e) 3.5 cm(a) 3 cm (b) 4 cm (c) 4.5 cm (d) 5 cm (e) 3.5 cm

If he looks from a point just above the glass slab, the ink mark will appear to be at 3/n = 3/(3/2) = 2 cm. Since he is looking from a distance of 5-3= 2 cm from the top of the slab, the ink mark will appear to be at 2 + 2 = 4cm.

Have you ever seen the setting sun while you are under water (while swimming)? Here is a question:

Have you ever seen the setting sun while you are under water (while swimming)? Here is a question:

**A diver under water sees the setting sun at an angle of nearly**

(a) 41º with respect to the horizontal (b) 49º with respect to the horizontal (c) 42º with the vertical (d) 45º with the horizontal (e) 37º with the horizontal.

The correct option is (a) since the critical angle for water is nearly 49º and this angle is with respect to the vertical in the present case. The angle with the horizontal is therefore 41º.(a) 41º with respect to the horizontal (b) 49º with respect to the horizontal (c) 42º with the vertical (d) 45º with the horizontal (e) 37º with the horizontal.

It will be convenient if you remember the critical angle of water and glass. In many problems you will encounter crown glass of refractive index 1.5. Its critical angle is 41.8º, which is roughly 42º. Critical angle for water is 48.75º, which is roughly 49º. You will come across many questions involving refractive index value of √2. This value has sanctity only for setting questions! It will be convenient to remember the critical angle of 45º for a substance having refractive index of √2. [Note that critical angle and refractive index of any medium is with respect to free space (or, air) unless specified otherwise].

Now, note this question which has been appearing in many test papers:

Now consider the following MCQ involving the minimum deviation produced by a prism:

When light rays are incident at an angle of 60º on one face of a glass prism, minimum deviation occurs. If the angle of minimum deviation also is 60º, what is the refractive index of the material of the prism?

(a) √3 (b) √3/2 (c) 1.5 (d) √2 (e) 1.6

Under the condition of minimum deviation, the angle of incidence (i

**A plane glass slab is kept over various coloured letters. The letter which appears least raised is**

(a) blue (b) violet (c) green (d) red (e) all are equally raised

Basically, you are asked to state which colour will produce the least apparent shift. The apparent shift = (Real distance - apparent distance) = t – (t/n). Since the refractive index (n) is the minimum for red, the least apparent shift is for red. So, red letters will appear least raised. [Since the refractive index is the maximum for violet, letters of violet colour will appear most raised].(a) blue (b) violet (c) green (d) red (e) all are equally raised

Now consider the following MCQ involving the minimum deviation produced by a prism:

When light rays are incident at an angle of 60º on one face of a glass prism, minimum deviation occurs. If the angle of minimum deviation also is 60º, what is the refractive index of the material of the prism?

(a) √3 (b) √3/2 (c) 1.5 (d) √2 (e) 1.6

Under the condition of minimum deviation, the angle of incidence (i

_{1}) and the angle of emergence (i_{2}) are equal (each equal to i). Since i_{1}+i_{2}-A = d**where‘d’ is the deviation, we have 2i – A = D when the deviation is the minimum (D). Therefore, 120º - A = 60º, from which A = 60º. The refractive index is given by n = sin [( A+D)/2] / sin (A/2) = (sin60)/sin30 = (√3 /2)/(1/2) = √3.**
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