Optics Question 896

Question: In YDSE a light containing two wavelengths 500 nm and 700 nm are used. Find the minimum distance where maxima of two wavelengths coincide. Given $ D/d=10^{3} $ , where D is the distance between the slits and the screen and d is the distance between the slits.

Options:

A) 1.2 m

B) 3.5 mm

C) 2.8 cm

D) 8.1 mm

Show Answer

Answer:

Correct Answer: B

Solution:

[b] At the place where maxima for both the wavelengths coincide, y will be same for both the maxima, i.e.,

$ \frac{n _{1}{\lambda _{1}}D}{d}=\frac{n _{2}{\lambda _{2}}D}{d}\Rightarrow \frac{n _{1}}{n _{2}}=\frac{{\lambda _{1}}}{{\lambda _{2}}}=\frac{700}{500}=\frac{7}{5} $

Minimum integral value of $ n _{2} $ is 5.

Minimum distance of maxima of the two Wave lengths fr 1 om central fringe $ \frac{n _{1}{\lambda _{2}}D}{d}=5\times 700\times {{10}^{-9}}\times 10^{3}=3.5mm. $



NCERT Chapter Video Solution

Dual Pane