1
MHT CET 2024 9th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Two wavelength 590 nm and 596 nm of sodium light are used one after other, to study the diffraction taking place at a single slit of aperture 2.4 mm . The distance between the slit and screen is 2 m . The separation between the positions of first secondary maximum of the diffraction pattern obtained in the two cases is

A
$7.5 \times 10^{-6} \mathrm{~m}$
B
$7.5 \times 10^{-9} \mathrm{~m}$
C
$ 2.5 \times 10^{-6} \mathrm{~m}$
D
$5.0 \times 10^{-6} \mathrm{~m}$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A parallel beam of light of intensity $I_0$ is incident on a glass plate, $25 \%$ of light is reflected by upper surface and $50 \%$ of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is

A
$\left[\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]^2$
B
$\left[\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]^2$
C
$\frac{5}{8}$
D
$\frac{8}{5}$
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A single slit of width $d$ is illuminated by violet light of wavelength 400 nm and the width of the diffraction pattern is measured as ' Y '. When half of the slit width is covered and illuminated by yellow light of wavelength 600 nm , the width of the diffraction pattern is

A
zero
B
$\frac{Y}{3}$
C
3 Y
D
4 Y
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In a biprism experiment, monochromatic light of wavelength ' $\gamma$ ' is used. The distance between the two coherent sources ' $d$ ' is kept constant. If the distance between slit and eyepiece ' $D$ ' is varied as $D_1, D_2, D_3, D_4$ and corresponding measured fringe widths are $\mathrm{W}_1, \mathrm{~W}_2, \mathrm{~W}_3, \mathrm{~W}_4$ then

A
$\mathrm{W}_1 \mathrm{D}_1=\mathrm{W}_2 \mathrm{D}_2=\mathrm{W}_3 \mathrm{D}_3=\mathrm{W}_4 \mathrm{D}_4$
B
$\frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}$
C
$\mathrm{W}_1 \sqrt{\mathrm{D}_1}=\mathrm{W}_2 \sqrt{\mathrm{D}_2}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}$
D
$\mathrm{D}_1 \sqrt{\mathrm{~W}_1}=\mathrm{D}_2 \sqrt{\mathrm{~W}_2}=\mathrm{D}_3 \sqrt{\mathrm{~W}_3}=\mathrm{D}_4 \sqrt{\mathrm{~W}_4}$
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