IP The figure(Figure 1) shows a single-slit diffraction pattern formed by light

Question

IP The figure(Figure 1) shows a single-slit diffraction pattern formed by light passing through a slit of width W= 14.2 m and illuminating a screen 0.885 m behind the slit.

Part A

What is the wavelength of the light?

=_______nm

Part B

If the width of the slit is decreased, will the distance indicated in the figure be greater than or less than 15.2 cm?

If the width of the slit is decreased, will the distance indicated in the figure be greater than or less than 15.2 ?

=_______nm

Part B

If the width of the slit is decreased, will the distance indicated in the figure be greater than or less than 15.2 cm?

Explanation / Answer

A)

L = slit to screen distance = 0.885 m

W = slit width = 14.2*10^-6 m

=Wavelength

y = the distance from the center of the central diffraction maximum to the first diffraction minimum

Now y = L /W =====> = yW/L

= 14.2*10^-6*y/0.885 = 16.05*y m (Put y to find wavelength)

b) As we decrease W, y will increase.

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