1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
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
4
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

Questions Asked from Ray & Wave Optics

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