Question

The width of the diffraction band varies

A. inversely as the wavelength
B. directly as the width of the slit
C. directly as the distance between the slit and the screen  
D. inversely as the size of the source from which the slit is illuminated
Answer :   directly as the distance between the slit and the screen
Solution :
The width of the diffraction band is given by
$$\eqalign{ & \beta = \frac{{\lambda D}}{d} \cr & \Rightarrow \beta \propto \lambda \cr & \beta \propto D\,\,{\text{and}}\,\,\beta \propto \frac{1}{d} \cr} $$

Releted MCQ Question on
Optics and Wave >> Wave Optics

Releted Question 1

In Young’s double-slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is

A. unchanged.
B. halved
C. doubled
D. quadrupled
Releted Question 2

Two coherent monochromatic light beams of intensities $$I$$ and $$4\,I$$  are superposed. The maximum and minimum possible intensities in the resulting beam are

A. $$5\,I$$  and $$I$$
B. $$5\,I$$  and $$3\,I$$
C. $$9\,I$$  and $$I$$
D. $$9\,I$$  and $$3\,I$$
Releted Question 3

A beam of light of wave length $$600\,nm$$  from a distance source falls on a single slit $$1mm$$  wide and a resulting diffraction pattern is observed on a screen $$2\,m$$  away. The distance between the first dark fringes on either side of central bright fringe is

A. $$1.2\,cm$$
B. $$1.2\,mm$$
C. $$2.4\,cm$$
D. $$2.4\,mm$$
Releted Question 4

Consider Fraunh offer diffraction pattern obtained with a single slit illuminated at normal incidence. At the angular position of the first diffraction minimum the phase difference (in radians) between the wavelets from the opposite edges of the slit is

A. $$\frac{\pi }{4}$$
B. $$\frac{\pi }{2}$$
C. $$2\,\pi $$
D. $$\pi $$

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