If $${I_0}$$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?
A.
$$4\,{I_0}$$
B.
$$2\,{I_0}$$
C.
$$\frac{{{I_0}}}{2}$$
D.
$${I_0}$$
Answer :
$$4\,{I_0}$$
Solution :
$$I = {I_0}{\left( {\frac{{\sin \phi }}{\phi }} \right)^2}\,\,{\text{and }}\phi = \frac{\pi }{\lambda }\left( {b\sin \theta } \right)$$
When the slit width is doubled, the amplitude of the wave at the centre of the screen is doubled, so the intensity at the centre is increased by a factor 4.
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
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 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
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