*I am neither especially clever nor especially gifted. I am only very, very curious*.

– Albert Einstein

The following question at the first glance may appear to be a difficult one to many of you; but, you will realise how easy it is when you apply basic points you studied in geometric optics:

A glass jar has the plane inner surface PQ of its bottom *silvered* and contains water (of refractive index *n* = 4/3) column of height *t* = 6 cm. A small light emitting diode (LED) is arranged at O at a height *d =* 8 cm from the water surface AB (Fig.). The silvered bottom of the jar acts as a plane mirror. At what distance from the free surface (AB) of water will this plane mirror form the image of the light emitting diode?

(a) 11 cm

(b) 14 cm

(c) 17 cm

(d) 18 cm

When you look into the plane mirror (silvered surface) PQ from the position O of the LED, the plane mirror will appear to be located at P_{1}Q_{1} (fig.) at a distance *t/n* from the free surface of water (because of normal refraction at the water surface). The distance of the LED from this refracted image P_{1}Q_{1 }of the plane mirror is therefore equal to (*d + t/n*) as shown in the adjoining figure.

The image of the LED must be formed at O_{1} which is at the same distance (*d + t/n*) from the *effective* plane mirror P_{1}Q_{1}.

As is clear from the adjoining figure, the distance of the image O_{1} from the free surface (AB) of water is (*d + t/n*) + *t/n* which is equal to (*d + *2*t/n*) = 8 + 2×6/(4/3) = 17 cm.

The following question appeared in EAMCET (Engineering) 2003 question paper:

One of the refracting surfaces of a prism of refractive index √2 is silvered. The angle of the prism is equal to the critical angle of a medium of refractive index 2. A ray of light incident on the unsilvered surface passes through the prism and retraces its path after reflection at the silvered face. Then the angle of incidence on the unsilvered surface is

(a) 0º

(b) 30º

(c) 45º

(d) 60º

The angle A of the prism (as mentioned in the question) is given by *n = *1/sin A where *n = *2.

[Remember *n = *1/sin *C* where *n* is the refractive index and* C* is the critical angle].

Therefore, sin A = ½ so that A = 30º

Since the ray retraces its path after reflection at the silvered face, it is incident *normally* at the silvered face (at the point N in the figure). With reference to the figure, angle QNA in the triangle QNA is 90º.

Since the angle A is 30º it follows that angle AQN = 60º so that the angle of refraction (*r*) at Q is 30º.

The angle of incidence (*i*) at the unsilvered face is given by

*n *= sin* i/*sin *r* from which sin *i = n *sin *r* = √2 sin 30º.

This gives sin *i =*1/√2 so that ** i = 45º**.

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