COMEDK 2024 Evening Shift | Wave Optics Question 6 | Physics

Incident light of wavelength $$\lambda=800 \mathrm{~nm}$$ produces a diffraction pattern on a screen $$1.5 \mathrm{~m}$$ away when it passes through a single slit of width $$0.5 \mathrm{~mm}$$. The distance between the first dark fringes on either side of the central bright fringe is

2.4 mm

2.4 cm

4.8 cm

4.8 mm

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

A slit of width $$10 \times 10^{-7} \mathrm{~m}$$ is illuminated by light of wavelength $$500 \mathrm{~nm}$$. Angular position of the first minimum is

$$\frac{1}{2}^0$$

30$$^\circ$$

1$$^\circ$$

60$$^\circ$$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

In Young's double slit experiment the ratio of phase difference between light waves reaching the third bright fringe and third dark fringe is

$$\frac{4}{3}$$

$$\frac{5}{2}$$

$$\frac{6}{5}$$

$$\frac{7}{6}$$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

In Young's double slit experiment, the ratio of intensities of light from one slit to the other is $$9: 1$$. If Im is the maximum intensity, what is the resultant intensity when they interfere at phase difference $$\phi$$ ?

$$ \frac{\operatorname{Im}}{9}\left[1+8 \cos ^2\left(\frac{\phi}{2}\right)\right] $$

$$ \frac{\operatorname{Im}}{4}\left[1+8 \cos ^2\left(\frac{\phi}{2}\right)\right] $$

$$ \frac{\operatorname{Im}}{4}\left[1+3 \cos ^2\left(\frac{\phi}{2}\right)\right] $$

$$ \frac{\operatorname{Im}}{2}\left[4+12 \cos ^2\left(\frac{\phi}{2}\right)\right] $$

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