(B) The formula for the position of the $n^{th}$ bright fringe (maxima) in Young's double slit experiment is given by $x_n = \frac{n \lambda D}{d}$.Gi

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(B) The formula for the position of the $n^{th}$ bright fringe (maxima) in Young's double slit experiment is given by $x_n = \frac{n \lambda D}{d}$.Given values are:$\lambda = 5000\;\mathring{A} = 5000 \times 10^{-10}\;m = 5 \times 10^{-7}\;m$$d = 0.2\;mm = 0.2 \times 10^{-3}\;m = 2 \times 10^{-4}\;m$$D = 200\;cm = 2\;m$For the third maximum, $n = 3$.Substituting these values into the formula:$x_3 = \frac{3 \times (5 \times 10^{-7}\;m) \times (2\;m)}{2 \times 10^{-4}\;m}$$x_3 = \frac{30 \times 10^{-7}}{2 \times 10^{-4}}\;m = 15 \times 10^{-3}\;m = 1.5 \times 10^{-2}\;m = 1.5\;cm$.Therefore, the third maximum is at $x = 1.5\;cm$.

Class 12 Physics Wave Optics Young's Double Slit Experiment (YDSE)

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In Young's double slit experiment, carried out with light of wavelength $\lambda = 5000\;\mathring{A}$, the distance between the slits is $d = 0.2\;mm$ and the screen is at $D = 200\;cm$ from the slits. The central maximum is at $x = 0$. The third maximum (taking the central maximum as the zeroth maximum) will be at $x$ equal to......$cm$.

AIPMT 1992 All AIPMT

A
$1.67$
B
$1.5$
C
$0.5$
D
$5$

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Wave Optics

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Young's Double Slit Experiment (YDSE)

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AIPMT 1992 Paper All AIPMT years →

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

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In Young's double-slit experiment,a monochromatic light source is used to illuminate slits $S_1$ and $S_2$. An interference pattern is obtained on the screen with a fringe width of $w$. If a mica sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits,the new fringe width $w'$ becomes:

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Two coherent sources $O_1$ and $O_2$ in Young's double slit experiment are illuminated with monochromatic light of wavelength $5000 \text{ Å}$. If a second order dark fringe is formed at a point $R$ on the screen,the path difference $O_1 R \sim O_2 R$ is (in $\mu m$)

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In an experiment, light passing through two slits separated by a distance of $0.3 \,mm$ is projected onto a screen placed at $1 \,m$ from the plane of the slits. It is observed that the distance between the central fringe and the adjacent bright fringe is $1.9 \,mm$. The wavelength of light in $nm$ is

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In Young's double slit experiment,the two slits act as coherent sources of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same setup,the two slits are of equal amplitude $A$ and wavelength $\lambda$ but are incoherent. The ratio of the intensity of light at the mid-point of the screen in the first case to that in the second case is

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Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)

Two coherent sources $O_1$ and $O_2$ in Young's double slit experiment are illuminated with monochromatic light of wavelength $5000 \text{ Å}$. If a second order dark fringe is formed at a point $R$ on the screen,the path difference $O_1 R \sim O_2 R$ is (in $\mu m$)

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