In Young's double-slit experiment,if $L$ is the distance between the slits and the screen upon which the interference pattern is observed,$x$ is the average distance between adjacent fringes,and $d$ is the slit separation,then the wavelength of light is given by:
- A$\frac{xd}{L}$
- B$\frac{xL}{d}$
- C$\frac{Ld}{x}$
- D$\frac{1}{Ldx}$
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In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?
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View SolutionLight of wavelength $6000 \, Å$ is incident on two slits. The distance between the slits is $0.1 \, cm$ and the screen is placed at a distance of $1 \, m$ from them. Find the angular position of the $10^{th}$ maximum in radians.
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View SolutionIn an interference experiment, the third bright fringe is obtained at a point on the screen with light of wavelength $700 \, nm$. What should be the wavelength of the light source in order to obtain the $5^{th}$ bright fringe at the same point (in $nm$)?
Two coherent sources $P$ and $Q$ produce interference at point $A$ on the screen,where a dark band is formed between the $4^{\text{th}}$ bright band and the $5^{\text{th}}$ bright band. The wavelength of the light used is $6000 \text{ Å}$. The path difference between $PA$ and $QA$ is:
In Young's double-slit experiment,which of the following phenomena occur?
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