The first minima due to a single slit diffraction is at -30- for a light of wavelength 5000 A- The width of slit is A- 5 - 10-5 cmB- 2-5 - 10-5 cmC- 10 - 10-5 cmD- 1-25 - 10-5 cm

The correct option is C 10 × 10^ 5cmSingle sit diffraction nbsp; sin θ = λ/a sin nbsp; 30^∘ = 5000 ∘A/a 1/2 = 5000 ∘A/a⇒ a = 10.000 ∘A a = 10 × 10^ 5 cm ...

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Standard XII

Physics

Refraction at a Spherical Surface

The first min...

Question

The first minima due to a single slit diffraction is at $θ = 30^{\circ}$ for a light of wavelength $5000 \circ A$ . The width of slit is

5 \times 10^{- 5} c m

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10 \times 10^{- 5} c m

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2.5 \times 10^{- 5} c m

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1.25 \times 10^{+ 5} c m

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Solution

The correct option is B $10 \times 10^{- 5} c m$
Single sit diffraction
$s i n θ = \frac{λ}{a} s i n 30^{\circ} = \frac{5000 \circ A}{a} \frac{1}{2} = \frac{5000 \circ A}{a} \Rightarrow a = 10.000 \circ A a = 10 \times 10^{- 5} c m$

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The first minima due to a single slit diffraction is at θ=30∘ for a light of wavelength 5000∘A. The width of slit is

The correct option is B 10×10−5cmSingle sit diffraction sinθ=λasin30∘=5000∘Aa12=5000∘Aa⇒a=10.000∘Aa=10×10−5cm

The first minima due to a single slit diffraction is at $θ = 30^{\circ}$ for a light of wavelength $5000 \circ A$ . The width of slit is

The correct option is B $10 \times 10^{- 5} c m$
Single sit diffraction
$s i n θ = \frac{λ}{a} s i n 30^{\circ} = \frac{5000 \circ A}{a} \frac{1}{2} = \frac{5000 \circ A}{a} \Rightarrow a = 10.000 \circ A a = 10 \times 10^{- 5} c m$