1
MCQ (Single Correct Answer)

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
2
MCQ (Single Correct Answer)

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
3
MCQ (Single Correct Answer)

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
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15cm from a converging lens of magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final image formed is:
A
real and at a distance of 6 cm from the convergent lens.
B
real and at a distance of 40 cm from convergent lens.
C
virtual and at a distance of 40 cm from convergent lens.
D
real and at a distance of 40 cm from the divergent lens.

Explanation

As parallel beam incident on diverging lens so the image will be formed at the focus of diverging lens.

$$ \therefore $$ v = –25 cm
The image formed by diverging lens is used as an object for converging lens.

So for converging lens

u = –25 – 15 = –40 cm,

f = 20 cm

So Final image formed by converging lens

$${1 \over {20}} = {1 \over V} - {1 \over { - 40}}$$

$$ \Rightarrow $$ V = 40 cm

V is positive so image will be real and will form at right side of converging lens at 40 cm.

Questions Asked from Ray & Wave Optics

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