Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 mathringA. When the slit is illuminated by light of another wavelength, the angular width decreases by 30 %. The wavelength of this light will be

Question

Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 mathringA. When the slit is illuminated by light of another wavelength, the angular width decreases by 30 %. The wavelength of this light will be (A) 6000 mathringA (B) 4200 mathringA (C) 3000 mathringA (D) 1800 mathringA

Tardigrade Admin · Accepted Answer

Angular width β=(2 λ/d) ⇒ (β1/β2)=(λ1/λ2) ⇒ (β/(70)100 β)=(6000/λ2) ⇒ λ2=4200 mathringA

$6000 \overset{˚}{A}$

$4200 \overset{˚}{A}$

$3000 \overset{˚}{A}$

$1800 \overset{˚}{A}$

Angular width $β = \frac{2 λ}{d}$
$\Rightarrow \frac{β _{1}}{β _{2}} = \frac{λ _{1}}{λ _{2}}$
$\Rightarrow \frac{β}{\frac{70}{100} β} = \frac{6000}{λ _{2}}$
$\Rightarrow λ_{2} = 4200 \overset{˚}{A}$

Q. Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000A˚. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will be

6000A˚

4200A˚

3000A˚

1800A˚

Angular width β=d2λ​ ⇒β2​β1​​=λ2​λ1​​ ⇒10070​ββ​=λ2​6000​ ⇒λ2​=4200A˚

$6000 \overset{˚}{A}$

$4200 \overset{˚}{A}$

$3000 \overset{˚}{A}$

$1800 \overset{˚}{A}$

Angular width $β = \frac{2 λ}{d}$
$\Rightarrow \frac{β _{1}}{β _{2}} = \frac{λ _{1}}{λ _{2}}$
$\Rightarrow \frac{β}{\frac{70}{100} β} = \frac{6000}{λ _{2}}$
$\Rightarrow λ_{2} = 4200 \overset{˚}{A}$