Question
A Young's double slit experiment is performed with white light:

Answer

  1. The central fringe will be white.
  2. There will not be a completely dark fringe.
  1. The fringe next to the central will be violet.
Explanation:
The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The output of the given circuit in Fig.
Curie temperature is the temperature above which:
If a spherical conductor comes out from the closed surface of the sphere then total flux emitted from the surface will be
The binding energy per nucleon of iron atom is approximately.
Which of the following gases has maximum $\text{rms}$ speed at a given temperature?
A magnetic field exerts no force on:
A charged particle goes undeflected in a region containing an electric and a magnetic field. It is possible that
  1. $\overrightarrow{\text{E}}||\overrightarrow{\text{B}},\ \overrightarrow{\text{v}}||\overrightarrow{\text{E}}$
  2. $\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}}$
  3. $\overrightarrow{\text{v}}||\overrightarrow{\text{B}}$ but $\overrightarrow{\text{E}}$ is not parallel to $\overrightarrow{\text{B}}$
  4. $\overrightarrow{\text{E}}||\overrightarrow{\text{B}}$ but $\overrightarrow{\text{v}}$ is not parallel to $\overrightarrow{\text{E}}$
What is the equivalent capacitance between  $A$ and $B$ in the given figure $($all are in farad$)$
If nearly $10^5$ coulomb liberate $1 \ gm$ equivalent of aluminium, then the amount of aluminium $($equivalent weight $9)$ deposited through electrolysis in $20$ minutes by a current of $50 \ \text{amp}$ will be
A bar magnet of magnetic moment $10^4 \mathrm{J} / \mathrm{T}$ is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of $4 \times$ $10^{-5} T$ to a direction $60^{\circ}$ from the field will be