1

### 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.
2

### 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
3

### 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
4

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

A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface. A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :

A
14 cm below flat surface
B
30 cm below flat surface
C
20 cm below flat surface
D
16 cm below flat surface

## Explanation

Using mirror formula.

${1 \over v} + {1 \over { - 4}} = {1 \over { - 5}}$

$\Rightarrow \,\,\,{1 \over v} = {1 \over 4} - {1 \over 5} = {1 \over {20}}$

$\Rightarrow \,\,\,v = 20\,$ cm

$\therefore$   Image distance is 20 cm

I1 acts as object for plane glass surface.

$\therefore$   Appartent depth =  ${{R + v} \over \mu } = {{30} \over {1.5}} = 20$ cm

$\therefore$   Position of the image of the air bubble is 20 cm below the flat surface.