6. ⇒ (MHT CET 2023 9th May Morning Shift )

In a single slit experiment, the width of the slit is doubled. Which one of the following statements is correct?

A. The intensity and width of the central maximum are unaffected.

B. The intensity remains same and angular width becomes half.

C. The intensity and angular width both are doubled.

D. The intensity increases by a factor 4 and the angular width decreases by a factor of $\frac{1}{2}$ .

Correct Option is (D)

For single slit diffraction,

Fringe width $W = \frac{λ D}{d}$

Also, $I \propto D^{2}$

$I \propto D^{2}$ This means I will be 4 times the original value of $I$ . As only 1 option mentions this, the correct answer is D.

7. ⇒ (MHT CET 2021 21th September Morning Shift )

In diffraction experiment, from a single slit, the angular width of central maximum does NOT depend upon

A. ratio of wavelength and slit width

B. distance of the slit from the screen

C. wavelength of light used

D. width of the slit

Correct Option is (B)

Angular width of central maximum $θ = \frac{2 λ}{a}$

Where a is slit width

8. ⇒ (MHT CET 2021 20th September Evening Shift )

In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5 mm$ . The screen on which the diffraction pattern is obtained is at a distance of $80 cm$ from the slit. The wavelength used is 6000 $\overset{o}{A}$ . The width of the silt is

A. 0.096 mm

B. 0.576 mm

C. 0.192 mm

D. 0.384 mm

Correct Option is (C)

$\begin{aligned} Distance between the first minima = \frac{2 λ D}{a} = 5 mm = 5 \times 10^{- 3} m \\ λ = 6000 \overset{o}{A} = 6 \times 10^{- 7} m, D = 80 cm = 0.8 m \\ ∴ \frac{2 \times 6 \times 10^{- 7} \times 0.8}{a} = 5 \times 10^{- 3} \\ ∴ a = \frac{2 \times 6 \times 10^{- 7} \times 0.8}{5 \times 10^{- 3}} = 0.192 \times 10^{- 3} m \\ = 0.192 mm \end{aligned}$

9. ⇒ (MHT CET 2021 20th September Morning Shift )

Light of wavelength ' $λ$ ' is incident on a single slit of width 'a' and the distance between slit and screen is 'D'. In diffraction pattern, if slit width is equal to the width of the central maximum then $D =$

A. $\frac{a^{2}}{λ}$

B. $\frac{a}{λ}$

C. $\frac{a^{2}}{2 λ}$

D. $\frac{a}{2 λ}$

Correct Option is (C)

Width of central maximum $= \frac{2 λ D}{a}$

$∴ a = \frac{2 λ D}{a}$

$∴ D = \frac{a^{2}}{2 λ}$

10. ⇒ (MHT CET 2021 20th September Morning Shift )

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 2m away from the lens. If wavelength of light used is 5000 $\overset{o}{A}$ then the distance between the first minimum on either side of the central maximum is ( $θ$ is small and measured in radian)

A. 2 $\times$ 10 $^{- 2}$ m

B. 10 $^{- 1}$ m

C. 10 $^{- 2}$ m

D. 10 $^{- 3}$ m

Correct Option is (D)

Distance between the first minima $= \frac{2 λ D}{a} = \frac{2 \times 5 \times 10^{- 7} \times 2}{0.2 \times 10^{- 3}} = 10^{- 3}$ m

⇒Wave Optics