1
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
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 $${{{1^o}} \over {60}}$$, the value of d0 is close to :
A
1 mm
B
2 mm
C
4 mm
D
3 mm
2
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
To find the focal length of a convex mirror, a student records the following data :

Object
Pin
Convex
Lens
Convex
Mirror
Image
Pin
22.2 cm 32.2 cm 45.8 cm 71.2 cm

The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :
A
f1 = 12.7 cm    f2 = 7.8 cm
B
f1 = 7.8 cm    f2 = 12.7 cm
C
f1 = 7.8 cm    f2 = 25.4 cm
D
f1 = 15.6 cm    f2 = 25.4 cm
3
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and aplane mirror are arranged as shown. For an object kept at a distance of 60 cm from the convex lens, the final image, formed by the combination, is a real image, at a distance of :

A
60 cm from the convex lens
B
60 cm from the concave lens
C
70 cm from the convex lens
D
70 cm from the concave lens
4
JEE Main 2016 (Offline)
+4
-1
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 $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$
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