JEE Main Result 2025 Session 2 Paper 1 (Out) - Paper 2 Scorecard Soon at jeemain.nta.nic.in

Lloyd's Mirror Experiment - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 3 Questions around this concept.

Solve by difficulty

A long narrow horizontal slit is placed 1mm above a horizontal plane mirror. The interference between the light coming directly from the silt and that after reflection is seen on a screen 1m away from the silt  if $\lambda$=700nm then fringe width is :

According to Lloyd's mirror experiment path difference for maximum intensity is

Concepts Covered - 1

Lloyd's mirror experiment

Lloyd's mirror experiment

In the Lloyd's mirror experiment, light from a monochromatic slit source reflects from a glass surface at a small angle and appears to come from a virtual source as a result. The reflected light interferes with the direct light from the source, forming interference fringes.

Experimental setup: 

A plane glass plate (acting as a miror) is illuminated at almost grazing incidence by a light from a sliftual image $S_2$ of $S_1$ is formed closed to $S_1$ by reflection and these two act as coherent sources. The expression giving the fringe width is the same as for
the double silt, but the fringe system differs in one important respect.

The path difference $S_2 P-S_1 P$ is a whole number of wavelengths, the fringe at P is dark not bright. This is due to $180^{\circ}$ phase change which occurs when light is reflected from a denser medium. At grazing incidence a fringe is formed at O , where the geometrical path difference between the direct and reflected waves is zero and it follows that it will be dark rather than bright.

Thus, whenever there exists a phase difference of a $\pi$ between the two interfering beams of light, conditions of maximas and minimas are interchanged, i.e.,
$\Delta x=n \lambda$ (for minimum intensity) and
$\Delta x=(2 n-1) \lambda / 2$ (for maximum intensity)

 

 

Study it with Videos

Lloyd's mirror experiment

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top