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Light of wavelength 600012 \times 10^{-6} \mathrm{~m}  is incident on a single slit. First minimum is obtained at a distance of 0.4 \mathrm{~cm}  from the centre. If width of the slit is 0.3 \mathrm{mm},  then distance between slit and screen will be

Option: 1

1.0 \mathrm{~m}


Option: 2

1.5 \mathrm{~m}


Option: 3

2.0 \mathrm{~m}


Option: 4

2.3 \mathrm{~m}


Answers (1)

best_answer

Using a \sin \theta=n \lambda

\text { or } a \cdot \frac{x}{D}=n \lambda \text { or } a=n \lambda \cdot \frac{D}{x}
\text { or } D=\frac{a \cdot x}{n \lambda}
\frac{3 \times 10^{-4} \times 4 \times 10^{-3}}{1 \times 6000 \times 10^{-10}} \mathrm{~m}=2.0 \mathrm{~m}

Posted by

Suraj Bhandari

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