QJEEM17Ph4633: JEE Main 2017 Physics - Class 12 JEEM17Ph4633 Solution: In a Young’s double slit experiment, slits are separated by 0.5 …

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QJEEM17Ph4633: JEE Main 2017 Physics - Class 12 JEEM17Ph4633 Solution: In a Young’s double slit experiment, slits are separated by 0.5 …

In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

Multiple Choice Options

A 15.6 mm
B 1.56 mm
C 7.8 mm
D 9.75 mm

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Accepted Answer

Let n₁^th fringe formed due to first wavelength and n₂^th fringe formed due to second wavelength coincide. So their distance from common central maxima will be same.

y_n₁ = y_n₂

\frac{n_{1} λ_{1} D}{d} = \frac{n_{2} λ_{2} D}{d}

\Rightarrow

\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}}

=

\frac{520 \times 10^{- 9}}{650 \times 10^{- 9}} = \frac{4}{5}

Hence, distance of the point of coincidence from the central maxima is

y =

\frac{n_{1} λ_{1} D}{d} = \frac{n_{2} λ_{2} D}{d}

=

\frac{4 \times 450 \times 10^{- 9} \times 15}{0.5 \times 10^{- 3}}

= 7.8 mm

QJEEM17Ph4633: JEE Main 2017 Physics - Class 12 JEEM17Ph4633 Solution: In a Young’s double slit experiment, slits are separated by 0.5 …

Multiple Choice Options

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