The maximum number of possible interference maxima when slit separation is equal to $4$ times the wavelength of light used in a double slit experiment is

\infty

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9

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8

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4

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Solution

The correct option is C $9$
$d s i n θ = n λ$
$s i n θ = \frac{n}{4}$
maximum values of $s i n θ$ can not exceed $\pm 1$
So possible value os n is $0, \pm 1, \pm 2, \pm 3, \pm 4$
answer is 9

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The maximum number of possible interference maxima when slit separation is equal to 4 times the wavelength of light used in a double slit experiment is

The correct option is C 9dsinθ=nλsinθ=n4maximum values of sinθ can not exceed ±1So possible value os n is 0,±1,±2,±3,±4answer is 9

The maximum number of possible interference maxima when slit separation is equal to $4$ times the wavelength of light used in a double slit experiment is

The correct option is C $9$
$d s i n θ = n λ$
$s i n θ = \frac{n}{4}$
maximum values of $s i n θ$ can not exceed $\pm 1$
So possible value os n is $0, \pm 1, \pm 2, \pm 3, \pm 4$
answer is 9