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Chapter 18: Problem 11

Monochromatic light from a narrow slit illuminates two parallel narrow slits producing an interference pattern on a screen. The separation between the two slits is now doubled and the distance between the screen and the slits is reduced to half. The fringe width (A) Is doubled (B) Becomes four times (C) Becomes one-fourth (D) Remains the same

Short Answer

Expert verified
The fringe width becomes one-fourth (Option C)

Step by step solution

01

Identify and Understand the Applicable Formula

The applicable formula for double-slit interference fringe width \( W \) is \( W = \frac{D \lambda}{d} \), where \( D \) is the distance to the screen, \( \lambda \) is the wavelength of light, and \( d \) is the separation of the slits.
02

Apply the Changes to the Variables in the Formula

Now, the separation between slits \( d \) is doubled, i.e., \( d \) becomes \( 2d \), the distance to the screen \( D \) is reduced to half, i.e., \( D \) becomes \( D/2 \)
03

Substitute the Changes into the Formula and Calculate

Substituting these new values into the formula gives a new fringe width \( W' = \frac{(D/2) \lambda}{2d} = \frac{D \lambda}{4d} \)
04

Compare New Fringe Width with Old Fringe Width

Comparing \( W' \) with the initial fringe width, \( W' = \frac{1}{4} W \), it is evident that the new fringe width \( W' \) becomes one-fourth of the original fringe width \( W \)

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