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Chapter 27: Q6PE (page 997)

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

Short Answer

Expert verified

The angle for the first-order maximum of the blue-light of 450-nm is 0.52掳.

Step by step solution

01

Given data

The wavelength of the blue light is 位=450nm10-9m1nm=4.5脳10-7m

The separation of the double slits is d=0.0500mm1m1000mm=5脳10-5m

02

Finding the angle which is the first-order maximum

A formula for the angle of the first-order maximum can be expressed as,

dsin胃=位鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌�.(1)

Substituting the given data in equation (1), we get,

\(\begin{array}{l}(5 \times {10^{ - 5}}\;m) \times \sin \theta = 4.5 \times {10^{ - 7}}\;m\\\theta = {\sin ^{ - 1}}\left( {\frac{{4.5 \times {{10}^{ - 7}}\;{\rm{m}}}}{{5 \times {{10}^{ - 5}}\;{\rm{m}}}}} \right)\\\theta = 0.52^\circ \end{array}\)

Therefore, the angle is0.52掳for the first-order maximum.

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Most popular questions from this chapter

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