In a Young’s double slit experiment, $I_{0}$ is the intensity at the central maximum and $β$ is the fringe width. The intensity at a point P distant x from the centre will be

Q. In a Young’s double slit experiment, let

I_{0}

be the intensity at the central maximum and

β

be the fringe width. Then intensity at a point

P

which is at a distance

x

from the central maximum is

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Youngs double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are βG, βR, βB and respectively. Then,

The correct option is D βR > βG > βBFring width, β=λDd and hence β∝λsince λred>λgreen>λblueSo βred>βgreen>βblueSo, the correct option will be (D)

Youngs double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are $β_{G}$ , $β_{R}$ , $β_{B}$ and respectively. Then,

The correct option is D $β_{R}$ > $β_{G}$ > $β_{B}$
Fring width, $β = \frac{λ D}{d}$ and hence $β \propto λ$
since $λ_{r e d} > λ_{g r e e n} > λ_{b l u e}$
So $β_{r e d} > β_{g r e e n} > β_{b l u e}$
So, the correct option will be $(D)$