Sunlight is used in a double-slit interference experiment. The fourth-order maxi

Question

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 470 nm occurs at an angle of = 90°. Thus, it is on the verge of being eliminated from the pattern because cannot exceed 90° in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change in separation is needed?

Explanation / Answer

For nth order maxima dsin = n where d is the slit separation.

For 4th order maxima =900 dsin90 = 4 = 4×470nm = 1880nm

d = 1880 nm

(a) For 3rd order maxima, 1880*sin = 3

sin = 3/1880

For > 626.66 nm , no value of is possible

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