JEE PYQ: Wave Optics Question 1
Question 1 - 2021 (16 Mar Shift 1)
A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is ‘$x$’ nm. The value of ‘$x$’ to the nearest integer is __________.
Show Answer
Answer: 600
Solution
$\beta = \frac{\lambda D}{d}$
$\lambda = \frac{\beta d}{D}$
$\lambda = \frac{6 \times 10^{-3} \times 10^{-3}}{10}$
$\lambda = 6 \times 10^{-7}$ m $= 600 \times 10^{-9}$ m
$\lambda = 600$ nm
BETA
