In a young's double slit experiment, the slits are placed 0.320mm apart. Light of wavelength λ = 500nm is incident on the slits. The total number of bright fringes that are observed in the angular range - 30 ^ o ≤ θ ≤

Correct answer: B. 641 dsinθ = nλ n = 0.320 × 10 ^ - 3 × (1)/(2) 500 × 10 ^ - 9 = 320 Hence number of maxima = 320 × 2 + 1 = 641

JEE Main 20199 JanuaryEvening Shift

JEE Main 2019 · 9 January · Evening Shift · Physics · Wave Optics

In a young's double slit experiment, the slits are placed 0.320mm...

In a young's double slit experiment, the slits are placed

0.320mm

apart. Light of wavelength

\lambda = 500nm

is incident on the slits. The total number of bright fringes that are observed in the angular range

- {30}^{o} \le \theta \le {30}^{o}

is:

B. 641

dsin\theta = n\lambda

n = \dfrac{0.320 \times {10}^{- 3} \times \dfrac{1}{2}}{500 \times {10}^{- 9}}

= 320

Hence number of maxima

= 320 \times 2 + 1 = 641