1
JEE Main 2020 (Online) 3rd September Evening Slot
+4
-1
Two light waves having the same wavelength $$\lambda$$ in vacuum are in phase initially. Then the first wave travels a path L1 through a medium of refractive index n1 while the second wave travels a path of length L2 through a medium of refractive index n2 . After this the phase difference between the two waves is :
A
$${{2\pi } \over \lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)$$
B
$${{2\pi } \over \lambda }\left( {{n_2}{L_1} - {n_1}{L_2}} \right)$$
C
$${{2\pi } \over \lambda }\left( {{{{L_1}} \over {{n_1}}} - {{{L_2}} \over {{n_2}}}} \right)$$
D
$${{2\pi } \over \lambda }\left( {{{{L_2}} \over {{n_1}}} - {{{L_1}} \over {{n_2}}}} \right)$$
2
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to
A
0.17o
B
1.7o
C
0.57o
D
0.07o
3
JEE Main 2020 (Online) 2nd September Evening Slot
+4
-1
In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be
A
28
B
24
C
30
D
18
4
JEE Main 2020 (Online) 2nd September Morning Slot
+4
-1
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
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