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

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0.01 rad

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1 rad

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0.1 rad

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Solution

The correct option is A
0.001 rad

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

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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 radUsing 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
$s i n θ = \frac{λ}{d} = \frac{550 \times 10^{- 9}}{0.55 \times 10^{- 3}} = 10^{- 3} = 0.001 r a d$