JEE Main 2021 (Online) 27th July Morning Shift | Wave Optics Question 86 | Physics

In the Young's double slit experiment, the distance between the slits varies in time as
d(t) = d₀ + a₀ sin$$\omega$$t; where d₀, $$\omega$$ and a₀ are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

$${{2\lambda D({d_0})} \over {(d_0^2 - a_0^2)}}$$

$${{2\lambda D{a_0}} \over {(d_0^2 - a_0^2)}}$$

$${{\lambda D} \over {d_0^2}}{a_0}$$

$${{\lambda D} \over {{d_0} + {a_0}}}$$

JEE Main 2021 (Online) 20th July Evening Shift

MCQ (Single Correct Answer)

-1

With what speed should a galaxy move outward with respect to earth so that the sodium-D line at wavelength 5890 $$\mathop A\limits^o $$ is observed at 5896 $$\mathop A\limits^o $$ ?

306 km/sec

322 km/sec

296 km/sec

336 km/sec

JEE Main 2021 (Online) 18th March Morning Shift

MCQ (Single Correct Answer)

-1

In Young's double slit arrangement, slits are separated by a gap of 0.5 mm, and the screen is placed at a distance of 0.5 m from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890 $$\mathop A\limits^o $$ is :-

1178 $$\times$$ 10^$$-$$9 m

1178 $$\times$$ 10^$$-$$6 m

1178 $$\times$$ 10^$$-$$12 m

5890 $$\times$$ 10^$$-$$7 m

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