In a single slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength λ '. The value of λ ' is

Question

In a single slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength λ '. The value of λ ' is (A) 4400 mathringA . (B) 6600 mathringA (C) 2000 mathringA (D) 3500 mathringA

Tardigrade Admin · Accepted Answer

In a single slit diffraction experiment, position of minima is given by

d s in θ = nλ

.
So, for first minima of red,

s in θ = 1 \times (\frac{λ _{R}}{d})

.
And as first maxima is midway between first and second minima, for wavelength

λ^{^{'}},

its position will be

d s in θ^{^{'}} = \frac{λ ^{^{'}} + 2 λ ^{^{'}}}{2} \Rightarrow s in θ^{^{'}} = \frac{3 λ ^{^{'}}}{2 d} .

According to given condition

s in θ = s in θ^{'}

\Rightarrow λ^{^{'}} = \frac{2}{3} λ_{R}

So,

λ^{^{'}} = \frac{2}{3} \times 6600 = 440 nm = 4400 \overset{˚}{A}

.

Q. In a single slit diffraction experiment, first minimum for red light $(660 nm)$ coincides with first maximum of some other wavelength $λ^{^{'}} .$ The value of $λ^{^{'}}$ is

$4400 \overset{˚}{A}$ .

$6600 \overset{˚}{A}$

$2000 \overset{˚}{A}$

$3500 \overset{˚}{A}$

Q. In a single slit diffraction experiment, first minimum for red light (660nm) coincides with first maximum of some other wavelength λ′. The value of λ′ is

4400A˚ .

6600A˚

2000A˚

3500A˚

Q. In a single slit diffraction experiment, first minimum for red light $(660 nm)$ coincides with first maximum of some other wavelength $λ^{^{'}} .$ The value of $λ^{^{'}}$ is

$4400 \overset{˚}{A}$ .

$6600 \overset{˚}{A}$

$2000 \overset{˚}{A}$

$3500 \overset{˚}{A}$