What will be the angular position for the first minimum due to Fraunhoffer diffraction with sources of light of wave length 550 nm and slit of width 0.55 mm

0.001 rad

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.01 rad

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 rad

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1 rad

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
0.001 rad

Using d sin $θ = n λ$ , for n =1
$s i n θ = \frac{λ}{d} = \frac{550 \times 10^{- 9}}{0.55 \times 10^{- 3}} = 10^{- 3} = 0.001 r a d$

Suggest Corrections

Similar questions

What will be the angular position for the first minimum due to Fraunhoffer diffraction with sources of light of wave length 550 nm and slit of width 0.55 mm

Q. What will be the angle of diffracting for the first minimum due to Fraunhoffer diffraction with sources of light of wave length 550 nm and slit of width 0.55 mm.

Q. The first diffraction minimum due to a single slit Fraunhoffer diffraction is at the angle of diffraction

30^{\circ}

for a light of wave length 5000

A^{0}

.The width of the slit is

Q. What will be the angle of diffraction for the first order maximum due to Fraunhofer diffraction by a single slit of width

0.50 m m

, using light of wavelength

500 n m

Q. What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength

6000 ˚ A

is used and slit width is

12 \times 10^{- 5} c m

Join BYJU'S Learning Program

What will be the angular position for the first minimum due to Fraunhoffer diffraction with sources of light of wave length 550 nm and slit of width 0.55 mm

The correct option is A 0.001 rad Using d sin θ=nλ, for n =1 sinθ=λd=550×10−90.55×10−3=10−3=0.001 rad

The correct option is A
0.001 rad

Using d sin $θ = n λ$ , for n =1
$s i n θ = \frac{λ}{d} = \frac{550 \times 10^{- 9}}{0.55 \times 10^{- 3}} = 10^{- 3} = 0.001 r a d$