In the post dated 6^{th} January 2008, the essentials of wave optics required for you to work out multiple choice questions were discussed. (You will find that post by clicking on the label ‘wave optics’ below this post). In that post, the following question was given for your practice:

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9** **

As promised, I give the solution here:

The ratio of intensities of light proceeding from the slits, I_{1}: I_{2} = 4 : 1

Therefore, the ratio of amplitudes of light proceeding from the slits,

a_{1}:a_{2} = √I_{1}: √I_{2} = 2:1

The amplitudes at the maximum and minimum of the interference pattern are therefore in the ratio (a_{1}+a_{2}) : (a_{1}–a_{2}) = 3:1

The intensities at the maximum and the minimum are in the ratio3^{2}:1^{2} = **9:1** [Option (e)].

[The intensity is proportional to the square of the amplitude].

We will discuss a few more typical multiple choice questions in this section:

**(1)** A single slit diffraction pattern is obtained on a screen using yellow light. If the yellow light is replaced by blue light without making any other changes in the experimental set up, what will happen to the diffraction bands?

(a) Bands will disappear

(b) Bands will become broader and farther apart

(c) Bands will become broader and crowded together

(d) Bands will become narrower and farther apart

(e) Bands will become narrower and crowded together** **

The angular width of the central maximum is **2 λ/a** and the minima are located at

*θ =*

*±*

*λ/a,*

*±*

**2**

*λ/a,*

*±*

**3**

*λ/a*…etc.This shows that when the wave length is *decreased* as is the case when *blue* light is used instead of yellow light, the bands will become narrower and crowded together** **[Option (e)].

**(2)** If white light is used in Young’s double slit experiment, what will happen to the interference bands?

(a) No bands will be obtained

(b) Many bands will be obtained as in the case of monochromatic light, but they will be coloured except the centre of the central band which will be white

(c) Very few bands will be obtained, but they will be coloured except the centre of the central band which will be white

(d) Many bands will be obtained as in the case of monochromatic light, but all of them will be white

(e)** **Very few bands will be obtained, but all of them will be white.

The centre of the central bright band (n = 0) will be white since all colours will satisfy the condition of zero path difference** **(and therefore** **of brightness). Since the condition for the first (order n =1) bright band is satisfied, at the point nearest to the centre, by the shortest wave length (violet), the first bright band will be coloured with the edge nearest to the centre *violet* and the edge farthest to the centre *red*. The remaining bright bands also will be coloured likewise. There will be very few distinct bands since the same point on the screen will satisfy the condition for brightness for one wave length and the condition for darkness for another.

The correct option therefore is (c).

**(3)** In an arrangement for Young’s double slit experiment, the separation between the slits is 1 mm. It is found that 8 bands of the double slit interference pattern can occupy the central maximum of the single slit diffraction pattern produced by one of the slits. What is the wjdth of each slit?

(a) 0.2 mm

(b) 0.25 mm

(c) 0.3 mm

(d) 0.35 mm

(e) 0.4 mm** **

The angular fringe width in the case of the double slit interference pattern is ** λ/d** so that the total angular separation of 8 interference bands is 8

*λ/d*. Since the angular width of the central maximum in the case of the single slit diffraction pattern is

**2**, we have

*λ/a* 8*λ/d* = 2*λ/a*.

Putting *d* = 1 mm, *a* = *d*/4 = 0.25 mm.

[You should note that in Young’s double slit experiment, the pattern you see on the screen is actually a superposition of single-slit diffraction from each slit, and the double-slit interference pattern. Therefore, there will be a broader diffraction peak in which several fringes of smaller width due to double-slit interference appear. The angular width of a double slit intetrference fringe is *λ/d *where as the angular width of the central maximum due to the single slit diffraction is 2*λ/a*. Therefore, the number of interference fringes occurring in the broad central diffraction maximum depends on the ratio *d*/*a*, that is the ratio of the distance between the two slits to the width of a slit. If ‘*a*’ is* *very small, the diffraction pattern will become very flat and we will observe the two-slit interference pattern with an appreciable number of interference maxima of almost equal intensity in the central region].

You will find some useful multiple choice questions in this section at physicsplus: Multiple Choice Questions on Wave Optics.

thanks . very good and honest effort

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