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Physics Ncert Solutions Class 12th Physics Class 12th Wave Optics

10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

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Payal Gupta | Contributor-Level 10

5 months ago

10.6 Wavelength of one light beam, $?_{1}$ = 650 nm

Wavelength of the other beam, $?_{2}$ = 520 nm

Distance of the screen from the slits = D

Distance between two slits = d

Distance of the $n^{t h}$ bright fringe on the screen from the central maximum is given by the relation,

$x$ = n $?_{1} (\frac{D}{d}$ )

For the 3rd bright fringe, n = 3

Hence $x$ = 3 $* 650 * (\frac{D}{d}$ ) nm

Let $n^{t h}$ bright fringe due to wave length $?_{2}$ and ${(n - 1)}^{t h}$ bright fringe due to wavelength $?_{1}$ coincide on the screen. We can equate the conditions for bright fringes as :

n $?_{2}$ = (n-1) $?_{1}$

520n = 650n – 650

n = $\frac{650}{130}$ = 5

The least distance

...more

10.6 Wavelength of one light beam, $?_{1}$ = 650 nm

Wavelength of the other beam, $?_{2}$ = 520 nm

Distance of the screen from the slits = D

Distance between two slits = d

Distance of the $n^{t h}$ bright fringe on the screen from the central maximum is given by the relation,

$x$ = n $?_{1} (\frac{D}{d}$ )

For the 3rd bright fringe, n = 3

Hence $x$ = 3 $* 650 * (\frac{D}{d}$ ) nm

Let $n^{t h}$ bright fringe due to wave length $?_{2}$ and ${(n - 1)}^{t h}$ bright fringe due to wavelength $?_{1}$ coincide on the screen. We can equate the conditions for bright fringes as :

n $?_{2}$ = (n-1) $?_{1}$

520n = 650n – 650

n = $\frac{650}{130}$ = 5

The least distance can be obtained from the relation,

$x$ = n $?_{2} (\frac{D}{d}$ ) = 5 $* 520 (\frac{D}{d}$ ) = 2600 $(\frac{D}{d}$ ) nm

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