Light of wavelength $589.3 nm$ is incident normally on the slit of width $0.1 mm$ . What will be the angular width of the central maximum, obtained on a screen placed at a distance of $1 m$ from the slit?

{0.67}^{\circ}

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{1.02}^{\circ}

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{1.34}^{\circ}

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None of the above

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Solution

The correct option is A ${0.67}^{\circ}$
Angular width of central maxima,

$= \frac{2 λ}{d} = \frac{2 \times 589.3 \times 10^{- 9}}{0.1 \times 10^{- 3}} in rad$

$= 0.0117 rad = 0.0117 \times \frac{180}{π} = {0.67}^{\circ}$

Hence, $(A)$ is the correct answer.

$Why this question?To understand the characteristics of diffraction pattern.$

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Light of wavelength 589.3 nm is incident normally on the slit of width 0.1 mm. What will be the angular width of the central maximum, obtained on a screen placed at a distance of 1 m from the slit?

The correct option is A 0.67∘Angular width of central maxima, =2λd=2×589.3×10−90.1×10−3 in rad =0.0117 rad=0.0117×180π=0.67∘ Hence, (A) is the correct answer. Why this question?To understand the characteristics of diffraction pattern.

Light of wavelength $589.3 nm$ is incident normally on the slit of width $0.1 mm$ . What will be the angular width of the central maximum, obtained on a screen placed at a distance of $1 m$ from the slit?