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Figure36-38gives αversus the sine of the angle θin a single-slit diffraction experiment using light of wavelength 610nm. The vertical axis scale is set by as role="math" localid="1663169810058" αs=12rad. What are (a) the slit width, (b) the total number of diffraction minima in the pattern (count them on both sides of the center of the diffraction pattern), (c) the least angle for a minimum, and (d) the greatest angle for a minimum?

Short Answer

Expert verified

(a) The slit width is 2330.03nm.

(b) The total number of diffraction minima in diffraction pattern is 4.

(c) The least angle for minimum is 15.2°.

(d) The greatest angle for minimum is 51.2°.

Step by step solution

01

Identification of given data

The wavelength of the light is λ=610nm.

The angle on vertical axis is αs=12rad.

02

Concept used

The angle of the diffraction is defined as the angle of central axis from the diffraction minima of different orders.

03

Determination of slit width

(a)

The slit width is given as:

αs=πdλ

Substitute all the values in equation.

12rad=πd610nmd=2330.03nm

Therefore, the slit width is 2330.03nm.

04

Determination of total number of diffraction minima in diffraction pattern

(b)

The total number of diffraction minima in diffraction pattern is given as:

mλ=dsinθ

Here, θis diffraction angle and its value for total number of minima is 90°. because beyond this angle diffraction is not observed.

Substitute all the values in equation.

m610nm=2330.03nmsin90°m=3.82m4

Therefore, the total number of diffraction minima in diffraction pattern is 4.

05

Determination of total number of diffraction minima in diffraction pattern

(c)

The least angle for minimum is given as:

sinθmin=mλd

The value of m is 1 for least angle for minimum.

Substitute all the values in equation.

sinθm=1610nm2330.03nmθmin=15.2°

Therefore, the least angle for minimum is 15.2°.

06

Determination of total number of diffraction minima in diffraction pattern

(d)

The greatest angle for minimum is given as:

sinθmax=mλd

The value of m is 3 for greatest angle for minimum.

Substitute all the values in equation.

sinθmax=3610nm2330.03nmθmax=51.2°

Therefore, the greatest angle for minimum is 51.2°.

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Most popular questions from this chapter

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui, Hawaii, and reflected back from the shuttle Discovery as it sped by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm. What is the effective diameter of the laser aperture at the Maui ground station? (Hint: A laser beam spreads only because of diffraction; assume a circular exit aperture.)

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