Subject

Physics

Class

CBSE Class 12

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Sample Papers

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 Multiple Choice QuestionsLong Answer Type

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31.

State the importance of coherent sources in the phenomenon of interference.

In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence, deduce the expression for the fringe width.

How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?


If coherent sources are not taken, the phase difference between two interfering waves, will change continuously and a sustained interference pattern will not be obtained. Thus, coherent sources provide sustained interference pattern.

Let S1 and S2 be two coherent sources separated by a distance d.

Let the distance of the screen from the coherent sources be D. Let point M be the foot of the perpendicular drawn from O, which is the midpoint of S1 and S2 on the screen. Obviously point M is equidistant from S1 and S2. Therefore the path difference between the two waves at point M is zero. Thus the point M has the maximum intensity.

Now, Consider a point P on the screen at a distance y from M. Draw S1N perpendicular from S1 on S2 P. 

 

The path difference between two waves reaching at P from S1 and S2 is given by, 

Error converting from MathML to accessible text. 

Expression for Fringe Width:

Distance between any two consecutive bright fringes or any two consecutive dark fringes are called the fringe width. It is denoted by straight omega . 

Let, y subscript n plus 1 end subscript space a n d space y subscript n be the distance of two consecutive fringes. Then, we have

y subscript n plus 1 end subscript space equals space left parenthesis n plus 1 right parenthesis fraction numerator D lambda over denominator d end fraction a n d space space y subscript n equals fraction numerator n D lambda over denominator d end fraction
So, fringe width = y subscript n plus 1 end subscript minus space space y subscript n space end subscript equals space left parenthesis n plus 1 right parenthesis fraction numerator D lambda over denominator d end fraction minus fraction numerator n D lambda over denominator d end fraction equals fraction numerator D lambda over denominator d end fraction

Fringe width is same for both bright and dark fringe.

When the entire apparatus is immersed in water, the fringe width decreases. 

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32.

(a) State Huygens principle. Using this principle explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally.

(b) Show that the angular width of the first diffraction fringe is half of that of the central fringe.

(c) If a monochromatic source of light is replaced by white light, what change would you observe in the diffraction pattern?

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