Optics Question 1104
Question: In a Young’s experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be
[CBSE PMT 1992; KCET 2004]
Options:
A) $ 60\times {{10}^{-4}}cm $
B) $ 10\times {{10}^{-4}}cm $
C) $ 10\times {{10}^{-5}}cm $
D) $ 6\times {{10}^{-5}}cm $
Show Answer
Answer:
Correct Answer: D
Solution:
Distance of $ n^{th} $ dark fringe from central fringe $ x _{n}=\frac{(2n-1)\lambda D}{2d} $
$ \therefore x _{2}=\frac{(2\times 2-1)\lambda D}{2d}=\frac{3\lambda D}{2d} $
$ \Rightarrow 1\times {{10}^{-3}}=\frac{3\times \lambda \times 1}{2\times 0.9\times {{10}^{-3}}}\Rightarrow \lambda =6\times {{10}^{-5}}cm $
BETA
