1
JEE Main 2017 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
In a Young’s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is
A
15.6 mm
B
1.56 mm
C
7.8 mm
D
9.75 mm
2
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600 nm. To see the stars just resolved by the telescope, the minimum distance between them should be (1 light year = 9.46 $$ \times $$ 1015 m) of the order of :
A
106 km
B
108 km
C
1011 km
D
1010 km
3
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In Young’s double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern. When the separation between the slits is varied, the interference pattern disappears for a particular distance d0 between the slits. If the angular resolution of the eye is $$({{{1}} \over {60}})^o$$, the value of d0 is close to :
A
1 mm
B
2 mm
C
4 mm
D
3 mm
4
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
The box of a pin hole camera, of length $$L,$$ has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $$\lambda $$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $${b_{\min }}$$) when :
A
$$a = \sqrt {\lambda L} \,$$ and $${b_{\min }} = \sqrt {4\lambda L} $$
B
$$a = {{{\lambda ^2}} \over L}$$ and $${b_{\min }} = \sqrt {4\lambda L} $$
C
$$a = {{{\lambda ^2}} \over L}$$ and $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$
D
$$a = \sqrt {\lambda L} $$ and $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$
JEE Main Subjects
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
Civil Services
UPSC Civil Service
CBSE
Class 12