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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°.

Step by step solution

01

Given data

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

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

02

Finding the angle which is the first-order maximum

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

dsinθ=λ………………………….(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°for the first-order maximum.

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Most popular questions from this chapter

Explain how microscopes can use wave optics to improve contrast and why this is important.

Using the result of the problem above, calculate the distance between fringes for 633nm light falling on double slits separated by 0.0800mm, located 3.00mfrom a screen as in Figure 27.56.

What type of experimental evidence indicates that light is a wave?

A non-reflective coating like the one described in Example 27.6 works ideally for a single wavelength and for perpendicular incidence. What happens for other wavelengths and other incident directions? Be specific.

(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?
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