In a single slit diffraction experiment, the first minimum for λ 1=600nm coincides with the first maximum for wavelength λ 2 . Calculate λ 2 .

Question

In a single slit diffraction experiment, the first minimum for λ 1=600nm coincides with the first maximum for wavelength λ 2 . Calculate λ 2 . (A) 400nm (B) 440nm (C) 600nm (D) 660nm

Tardigrade Admin · Accepted Answer

In the case of a single-slit experiment, for minima,

a s in θ = nλ

,
and for maxima,

a s in θ = \frac{( 2 n + 1 )}{2} λ

.
So, for the first minima and maxima,

\frac{a s in θ}{a s in θ} = \frac{1 ( λ ) _{1}}{( ( 3 ( λ ) _{2} ) /2 )} \Rightarrow \frac{3 ( λ ) _{2}}{2} = 600 nm

\Rightarrow λ_{2} = 400 nm

.

Q. In a single slit diffraction experiment, the first minimum for $λ_{1} = 600 nm$ coincides with the first maximum for wavelength $λ_{2}$ . Calculate $λ_{2}$ .

$400 nm$

$440 nm$

$600 nm$

$660 nm$

Q. In a single slit diffraction experiment, the first minimum for λ1​=600nm coincides with the first maximum for wavelength λ2​ . Calculate λ2​ .

400nm

440nm

600nm

660nm

In the case of a single-slit experiment, for minima, asinθ=nλ , and for maxima, asinθ=2(2n+1)​λ . So, for the first minima and maxima, asinθasinθ​=((3(λ)2​)/2)1(λ)1​​⇒23(λ)2​​=600nm ⇒λ2​=400nm .

Q. In a single slit diffraction experiment, the first minimum for $λ_{1} = 600 nm$ coincides with the first maximum for wavelength $λ_{2}$ . Calculate $λ_{2}$ .

$400 nm$

$440 nm$

$600 nm$

$660 nm$