A slit of width a is illuminated by parallel monochromatic light of wavelength - The value of a at which the first minimum of the diffraction pattern will from at - -30- is

For the first minima, asinθ=nλ Here, θ=30^∘ ; n=1 ⇒ a=λsin 30^∘=2λ lt;!–td border: 1px solid #ccc;br mso data placement:same cell;– gt; Hence, (C) ...

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A slit of wid...

Question

A slit of width $a$ is illuminated by parallel monochromatic light of wavelength $λ .$ The value of $a$ at which the first minimum of the diffraction pattern will from at $θ = 30^{\circ}$ is

\frac{λ}{2}

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λ

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2 λ

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3 λ

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Solution

The correct option is C $2 λ$
For the first minima,

$a sin θ = n λ$

Here, $θ = 30^{\circ}; n = 1$

$\Rightarrow a = \frac{λ}{sin 30^{\circ}} = 2 λ$

 Hence, $(C)$ is the correct answer.

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A slit of width a is illuminated by parallel monochromatic light of wavelength λ. The value of a at which the first minimum of the diffraction pattern will from at θ=30∘ is

The correct option is C 2λFor the first minima, asinθ=nλ Here, θ=30∘ ; n=1 ⇒ a=λsin30∘=2λ  Hence, (C) is the correct answer.

A slit of width $a$ is illuminated by parallel monochromatic light of wavelength $λ .$ The value of $a$ at which the first minimum of the diffraction pattern will from at $θ = 30^{\circ}$ is

The correct option is C $2 λ$
For the first minima,

$a sin θ = n λ$

Here, $θ = 30^{\circ}; n = 1$

$\Rightarrow a = \frac{λ}{sin 30^{\circ}} = 2 λ$

 Hence, $(C)$ is the correct answer.