Answer :

(a)

Consider two coherent sources of light S_{1} and S_{2}. The two sources of light will produce an interference pattern on the screen GG’. Consider a point P on the screen. Let be the phase difference due to the path difference S_{2}P - S_{1}P .

The electric filed from the light at each of the sources S_{1} and S_{2} can be written as,

where each source of light has maximum electric filed strength .

At P, by principle of superposition,

As intensity is the square of the amplitude,

Note that . Hence,

(b)

Given,

Size of aperture,

Wavelength of light,

Distance from screen,

Size of one maxima/minima (except the central),

Now, the required patriation has four fringes. Hence, the distance is:

OR

(a)

The conditions for total internal reflection are:

(i) Light must be trying to travel from optically denser medium to optically lighter medium.

(ii) The angle of incidence must be greater than a certain angle known as critical angle.

__Derivation:__

Suppose light is travelling from medium 1 to medium 2. When light strikes the surface separating the media, some of it refracts and some of it reflects. If we keep increasing the angle of incidence, the refracted ray becomes parallel to the surface. This value of angle of incidence is known as the critical angle. If we increase the angle of incidence, even further, we will have a reflected ray and no refracted ray. Thus, we will have total internal reflection.

Let us consider the condition when the angle of incidence is equal to critical angle, .

By Snell’s law,

where and are the refractive indices of medium 1 and medium 2 respectively. Let be the refractive index of 1 with respect to 2.

Thus, we have obtained the required relation.

(b)

**Concepts/Formulas used:**

__The lens equation:__

The lens formula is:

where f is the focal length, u is the object distance and v is the image distance. All quantities are measured with the cartesian sign convention (the sign convention in NCERT).

Let us consider the lenses one by one.

The first lens forms an image of the given object.

Here,

Object distance,

Focal length,

Using the lens equation,

The image I_{1} formed by the first lens acts as an object for the second lens.

Here,

Object distance,

Focal length,

Let us now consider the third lens. It is a converging lens as it has a positive focal point. An object at infinity would have its image at the focal point (either of the focal points as the object is in either direction).

Hence, the final image is formed at 30cm from the third lens at either side.

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