1
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

A double slit experiment is immersed in water of refractive index 1.33. The slit separationis 1 $$\mathrm{mm}$$ and the distance between slit and screen is $$1.33 \mathrm{~m}$$. The slits are illuminated by a light of wavelength $$6300\,\mathop A\limits^o $$. The fringewidth is

A
$$4.9 \times 10^{-4} \mathrm{~m}$$
B
$$6.3 \times 10^{-4} \mathrm{~m}$$
C
$$8.6 \times 10^{-4} \mathrm{~m}$$
D
$$5.8 \times 10^{-4} \mathrm{~m}$$
2
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

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

A
0.096 mm
B
0.576 mm
C
0.192 mm
D
0.384 mm
3
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

In Young's double slit experiment, with a source of light having wavelength $$6300 \mathop A\limits^o$$, the first maxima will occur when the

A
path difference is $$9200 \mathop A\limits^o$$
B
phase difference is $$\mathrm{n}$$ radian
C
phase difference is $$\frac{\pi}{2}$$ radian.
D
path difference is $$6300 \mathop A\limits^o$$
4
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

In Young's double slit experiment, the intensity at a point where the path difference is $$\frac{\lambda}{4}$$ [ $$\lambda$$ is wavelength of light used] is '$$\mathrm{I}$$'. If '$$\mathrm{I}_0$$' is the maximum intensity then $$\frac{\mathrm{I}}{\mathrm{I}_0}$$ is equal to $$\left[\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$$

A
3 : 2
B
2 : 3
C
3 : 4
D
1 : 2
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