1

### 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
2

### 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.
3

### JEE Main 2016 (Offline)

Arrange the following electromagnetic radiations per quantum in the order of increasing energy :

A : Blue light           B : Yellow light

C : X-ray                 D : Radiowave.
A
C, A, B, D
B
B, A, D, C
C
D, B, A, C
D
A, B, D, C
4

### JEE Main 2017 (Offline)

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 ×108 ms–1)
A
15.3 GHz
B
10.1 GHz
C
12.1 GHz
D
17.3 GHz

## Explanation

This question is from Doppler's effect of light.

When observer is moving towards the source then the frequency of wave measured by the observer will be

fobserved = factual$\sqrt {{{c + v} \over {c - v}}}$

where c = speed of light and v = speed of observer

According to the question, v = ${c \over 2}$

$\therefore$ fobserved = factual$\sqrt {{{c + {c \over 2}} \over {c - {c \over 2}}}}$

= 10$\times$$\sqrt {{{{{3c} \over 2}} \over {{c \over 2}}}}$

= 10$\sqrt 3$ = 17.3 GHz