question_answer

In YDSE screen is kept 0.8 m from the slits. The coherent sources are 0.016 cm apart & fringes are observed on the screen. It is found that with a certain monochromatic source of light, the fourth bright fringe is situated at a distance of 1.06 cm from the central fringe. The wavelength of light used is

A)  480nm                                

B)  530 nm  

C)  650 nm                               

D)  580 nm

Correct Answer: B

Solution :

For maxima, \[\Delta x=n\lambda \]\[\Rightarrow \]\[\frac{dy}{D}=n\lambda \] For fourth maxima, n = 4 \[\frac{dy}{D}=4\lambda \Rightarrow \lambda =\frac{dy}{4D}=\frac{0.016\times {{10}^{-2}}\times 1.06\times {{10}^{-2}}}{4\times 0.8}\] \[\Rightarrow \lambda =530nm\]