Question

In a YDSE bi-chromatic light of wavelengths, $400 \, nm$ and $560 \, nm$ are used. The distance between the plane of the slits is $0.1 \, mm$ and the distance between 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)

Let $n \text{th}$ minimum of $400\,nm$ coincides with $m \text{th}$ minimum of $560 \,nm$, then
For $n^{th}$ minima $\delta=\left(2 n - 1\right)\frac{\lambda }{2}$
$\left(2 n - 1\right) \left(\frac{4 0 0}{2}\right) = \left(2 \text{m} - 1\right) \left(\frac{5 6 0}{2}\right)$
$\frac{2 n - 1}{2 \text{m} - 1} = \frac{7}{5} = \frac{1 4}{1 0} = \frac{2 1}{1 5}$
i.e., $4 \text{th}$ minimum of $400 \,nm$ coindes with $3$rd minimum of $560 \,nm$.
Distance of $n^{th}$ minima from center $=\frac{\left(2 n - 1\right) \lambda D}{2 d}$
Location of this minimum is,
$\left(\text{Y}\right)_{1} = \frac{\left(2 \times 4 - 1\right) \left(1 0 0 0\right) \left(4 0 0 \times 1 0^{- 6}\right)}{2 \times 0 \cdot 1} = 1 4 \,\text{mm}$
Next $11$th minimum of $400 \,nm$ will coincide with $8$th
minimum of $560\, nm$.
Location of this minimum is,
$\left(\text{Y}\right)_{2} = \frac{\left(2 \times 1 1 - 1\right) \left(1 0 0 0\right) \left(4 0 0 \times 1 0^{- 6}\right)}{2 \times 0 \cdot 1} = 4 2 \, \text{mm}$
$ \, ∴ \, $ Required distance $= \text{Y}_{2} - \text{Y}_{1} = 2 8 \,\text{mm}$
Hence, the correct option is ($28 \,mm$)