1

### 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.
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### 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
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### 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
4

### JEE Main 2017 (Offline)

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

## Explanation

Let n1th fringe formed due to first wavelength and n2th fringe formed due to second wavelength coincide. So their distance from common central maxima will be same.

yn1 = yn2

${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$

$\Rightarrow$ ${{{n_1}} \over {{n_2}}} = {{{\lambda _2}} \over {{\lambda _1}}}$ = ${{520 \times {{10}^{ - 9}}} \over {650 \times {{10}^{ - 9}}}} = {4 \over 5}$

Hence, distance of the point of coincidence from the central maxima is

y = ${{{n_1}{\lambda _1}D} \over d} = {{{n_2}{\lambda _2}D} \over d}$ = ${{4 \times 450 \times {{10}^{ - 9}} \times 15} \over {0.5 \times {{10}^{ - 3}}}}$ = 7.8 mm