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What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two slits is 1 mm?

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#### Solution

Let the width of each slit be ‘a’.

The separation between m maxima in a double slit experiment is given by y_{m}.

`y_m=m(lambdaD)/d`

where D is the distance between the screen and the slit and d is the separation between the slits.

We know that the angular separation between m maxima can be given as

`theta_m=y_m/D=m((lambdaD)/d)/D`

`=>theta_m=(mlambda)/d`

Therefore, we can write the angular separation between 10 bright fringes as

`=>theta_10=(10lambda)/d" ....(1)"`

The angular width of the central maximum in the diffraction pattern due to a single slit of width ‘a’ is given by

`2theta_1=2(lambda/a)" ....2"`

It is given that 10 maxima of the double slit pattern is formed within the central maximum of the single slit pattern.

Therefore, we can equate (1) and (2) as

`(10lambda)/d=(2lambda)/a`

Solving, we get

`a=(2lambdad)/(10lambda)`

`a=d/5`

Given that the seperation between the slits = 1 mm

Therefore the slit width `a=d/5=(1mm)/5=0.2 mm`

Therefore, the width of each slit = 0.2 mm

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