1
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is :
A
71.6o
B
90o
C
18.4o
D
45o
2
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to :
A
22 mm
B
12 mm
C
33 mm
D
2 mm
3
JEE Main 2020 (Online) 7th January Morning Slot
+4
-1
Visible light of wavelength 6000 $$\times$$ 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at $$\theta$$1, then $$\theta$$1, is close to :
A
45o
B
30o
C
25o
D
20o
4
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
A transparent cube of side, made of a material of refractive index $$\mu$$2, is immersed in a liquid of refractive index $$\mu$$1($$\mu$$1 < $$\mu$$2). A ray is incident on the face AB at an angle $$\theta$$(shown in the figure). Total internal reflection takes place at point E on the face BC.
Then $$\theta$$ must satisfy :
A
$$\theta > {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1}$$
B
$$\theta < {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1}$$
C
$$\theta < {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}$$
D
$$\theta > {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}$$
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