First slide

Young's double slit Experiment

Question

In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m. The minimum distance between two successive regions of complete darkness is

Moderate

A

4 mm
B

5.6 mm
C

14 mm
D

28 mm

Solution

Let n^th minima of 400 nm coincides with m^th minima of 560 nm then
$(2 n - 1) 400 = (2 m - 1) 560$
$\Rightarrow \frac{2 n - 1}{2 m - 1} = \frac{7}{5} = \frac{14}{10} = \frac{21}{15}$
i.e., 4th minima of 400 nm coincides with 3rd minima of 560 nm.
The location of this minima is
$= \frac{7 (1000) (400 \times 10^{- 6})}{2 \times 0.1} = 14 mm$
Next, 11th minima of 400 nm will coincide with 8^th minima of 560 nm
Location of this minima is
$= \frac{21 (1000) (400 \times 10^{- 6})}{2 \times 0.1} = 42 mm$
$∴$ Required distance = 28 mm

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