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Chapter 33: Q. 3 (page 955)

A double-slit experiment is performed with light of wavelength630nm. The bright interference fringes are spaced 1.8mm apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 420nm ?

Short Answer

Expert verified

Due to light changes fringe spacing is1.2mm.

Step by step solution

01

Formula for position of bright fringes

In the double slit experiment, the position of the bright fringes can be written as follows:

ym=mλLd

As a result, the distance between any two consecutive brilliant fringes is

Δy=ym+1-ym=(m+1)λLd-mλLd=λLd

02

Calculation for fringe space

This equation can be rearranged to find the ratio (L/d).

Ld=Δyλ

=1.8×10-3m630×10-9m

=2857

This constant ratio can now be used to calculate the fringe spacing for a wavelength of 420nm.

Δy=λLd=420×10-9m×2857

Δy=1.2mm

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Most popular questions from this chapter

Optical computers require microscopic optical switches to turn signals on and off. One device for doing so, which can be implemented in an integrated circuit, is the Mach-Zender interferometer seen in FIGURE. Light from an on-chip infrared laser (λ=1.000μm)is split into two waves that travel equal distances around the arms of the interferometer. One arm passes through an electro-optic crystal, a transparent material that can change its index of refraction in response to an applied voltage. Suppose both arms are exactly the same length and the crystal’s index of refraction with no applied voltage is1.522.

a. With no voltage applied, is the output bright (switch closed, optical signal passing through) or dark (switch open, no signal)? Explain.

b. What is the first index of refraction of the electro-optic crystal larger than 1.522that changes the optical switch to the state opposite the state you found in part a?

In a double-slit interference experiment, which of the following actions (perhaps more than one) would cause the fringe spacing to increase? (a) Increasing the wavelength of the light. (b) Increasing the slit spacing. (c) Increasing the distance to the viewing screen. (d) Submerging the entire experiment in water.

You've found an unlabeled diffraction grating. Before you can use it, you need to know how many lines per it has. To find out, you illuminate the grating with light of several different wavelengths and then measure the distance between the two first-order bright fringes on a viewing screen 150cmbehind the grating. Your data are as follows:


Use the best-fit line of an appropriate graph to determine the number of lines per mm.

In a single-slit experiment, the slit width is 200 times the wavelength of the light. What is the width (inmm)of the central maximum on a screen 2.0m behind the slit?

Light of 630nmwavelength illuminates two slits that are0.25mmapart. FIGURE EX33.5shows the intensity pattern seen on a screen behind the slits. What is the distance to the screen?
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