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. $
BETA
