JEE MAIN - Physics (2017 (Offline) - No. 3)
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
15.6 mm
1.56 mm
7.8 mm
9.75 mm
설명
Let n1th fringe formed due to first wavelength and n2th fringe formed due to second wavelength coincide. So their distance from common central maxima will be same.
yn1 = yn2
$${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$$
$$ \Rightarrow $$ $${{{n_1}} \over {{n_2}}} = {{{\lambda _2}} \over {{\lambda _1}}}$$ = $${{520 \times {{10}^{ - 9}}} \over {650 \times {{10}^{ - 9}}}} = {4 \over 5}$$
Hence, distance of the point of coincidence from the central maxima is
y = $${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$$ = $${{4 \times 450 \times {{10}^{ - 9}} \times 15} \over {0.5 \times {{10}^{ - 3}}}}$$ = 7.8 mm
yn1 = yn2
$${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$$
$$ \Rightarrow $$ $${{{n_1}} \over {{n_2}}} = {{{\lambda _2}} \over {{\lambda _1}}}$$ = $${{520 \times {{10}^{ - 9}}} \over {650 \times {{10}^{ - 9}}}} = {4 \over 5}$$
Hence, distance of the point of coincidence from the central maxima is
y = $${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$$ = $${{4 \times 450 \times {{10}^{ - 9}} \times 15} \over {0.5 \times {{10}^{ - 3}}}}$$ = 7.8 mm