In a Young's double slit experiment, a parallel beam of monochromatic light is used. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. If the incident beam makes an

Question

Quantrex Academy · Accepted Answer

Correct answer: C. A dark fringe

The path difference introduced before the light reaches the slits is

\Delta x_1 = d \sin \theta

.

Given

\theta = \sin^{-1}

\left(\dfrac{\lambda}{2d}\right), we have

\sin \theta =

\dfrac{\lambda}{2d}.

Substituting the value of

\sin \theta

, we get

\Delta x_1 = d \left(\dfrac{\lambda}{2d}\right) = \dfrac{\lambda}{2}.<br><br>At the centre of the screen

P_0

, the path difference introduced after the light passes through the slits is

\Delta x_2 = 0

.<br><br>The total path difference at

P_0

is

\Delta x = \Delta x_1 + \Delta x_2 =

\dfrac{\lambda}{2} +

0 =

\dfrac{\lambda}{2}.<br><br>A path difference of \dfrac{\lambda}{2} corresponds to a phase difference of

\pi

, which results in destructive interference.<br><br>Therefore, a dark fringe will be formed at the centre

P_0

.

In a Young's double slit experiment, a parallel beam of...

Options

Correct answer

Step-by-step solution

More from Light Waves