1
JEE Main 2020 (Online) 2nd September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source ($$\lambda $$ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at a distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close is
A
1.27 $$\mu $$m
B
2.05 $$\mu $$m
C
2.87 nm
D
2 nm
2
JEE Main 2020 (Online) 8th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is $${1 \over 8}$$th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is :
A
0.853
B
0.568
C
0.672
D
0.760
3
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :
A
4.9 mm
B
5.9 mm
C
6.9 mm
D
3.9 mm
4
JEE Main 2020 (Online) 7th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
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