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

• A. 4 mm
• B. 5.6 mm
• C. 14 mm
• D. 28 mm

Answer 1

Answer: (D)

28 mm

Answer 2

Let nth minima of 400 nm coincides with mth minima of
560 nm, then
\hspace10mm (2n-1) \big(\frac{400}{2}\big)=(2m-1)\big(\frac{560}{2}\big)
or \hspace20mm \frac{2n-1}{2m-1}=\frac{7}{5}=\frac{14}{10}=...
i.e. 4th minima of 400 nm coincides with 3rd minima of
560 nm.

Location of this minima is,
$\, \, \, Y_1=\frac{(2 \times 4-1)(1000)(400 \times 10^{-6})}{2 \times 0.1}=14$ mm
Next 11th minima of 400 nm will coincide with 8 th minima
of 560 nm
Location of this minima is,
$\, \, \, Y_2=\frac{(2 \times 11-1)(1000)(400 \times 10^{-6})}{2 \times 0.1}=42$ mm
$\therefore$ Required distance =$Y_2-Y_1$=28 mm
Hence, the correct option is (d).