1

### JEE Main 2016 (Online) 9th April Morning Slot

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

## Explanation

We know,

Fringe width, $\beta$ = ${{\lambda D} \over {{d_0}}}$ From image,

$\theta$ = ${\beta \over D}$

$\Rightarrow$  $\theta$ = ${\lambda \over {{d_0}}}$

$\Rightarrow$  d0 = ${\lambda \over \theta }$

= ${{600 \times {{10}^{ - 9}}} \over {{1 \over {60}} \times {\pi \over {180}}}}$

= 2.06 $\times$ 10$-$3

$\simeq$ 2 mm
2

### JEE Main 2016 (Online) 9th April Morning Slot

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

## Explanation

For convex lens,

${1 \over {30}} = {1 \over {{V_1}}} + {1 \over {60}}$

$\Rightarrow$  ${1 \over {{V_1}}} = {1 \over {60}}$

$\Rightarrow$  ${V_1} = 60$ cm

For concave lens,

${1 \over { - 120}} = {1 \over {{V_2}}} - {1 \over {40}}$

$\Rightarrow$  ${1 \over {{V_2}}} = {1 \over {60}}$

$\Rightarrow$  ${V_2} = 60$ cm

object will be at 60 cm distance from concave lens.

As mirror distance is 50 cm from concave lens. So virtual object is 10 cm behind mirror.

So, real image is 10 cm infront of mirror.

Distance of image from concave lens $=$ 50 $-$ 10 $=$ 40 cm. Which is same for the real object.

So similar situation will happen and final image will form at the real object it self.
3

### JEE Main 2016 (Online) 10th April Morning Slot

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

## Explanation

The limit of resolution of a telescope,

$\Delta$$\theta$  =  ${{1.22\,\,\lambda } \over D}$ = ${l \over R}$

$\therefore$   $l$ = ${{1.22\,\,\lambda R} \over D}$

=   ${{1.22 \times 6 \times {{10}^{ - 7}} \times 10 \times 9.46 \times 10{}^{15}} \over {30 \times {{10}^{ - 2}}}}$

=   2.31 $\times$ 108 km
4

### JEE Main 2016 (Online) 10th April Morning Slot

To determine refractive index of glass slab using a travelling microscope, minimum number of readings required are :
A
Two
B
Three
C
Four
D
Five