DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE

JEE: DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE

The document DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE is a part of the JEE Course DC Pandey Solutions for NEET Physics.
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Introductory Exercise 27.1

Ques 1: Show that the unit of DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE is m/s.
Sol:
 DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE = Speed of light in vacuum.

Ques 2: The magnetic field in a plane electromagnetic wave is given by (SI units)
By = (2 × 10-7 T) sin [500 x + 1.5 × 1011 t]
(a) What is the wavelength and frequency of the wave?
(b) Write an expression for the electric field vector.
Ans:
 DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
= 0.0125 m= 1.25 cm
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
= 2.39 × 1010Hz
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
E0 = cB0
= (3 × 108)(2 × 10-7) = 60 V/m
kx and ωt have same sign. Therefore, electromagnetic wave is travelling along negative x direction. DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE are in mutually perpendicular direct ions. Hence oscillations DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE of are along z-direction.

Introductory Exercise 27.2

Ques 1: Prove that for any value of angle i, rays 1 and 2 are parallel.
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Sol: 
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
PQ and MN are mutually parallel. Rays 1 and 2 are making equal angles (= ∠i) from PQ and MN. So, they are mutually parallel.

Ques 2: A man approaches a vertical plane mirror at speed of 2 m/s. At what rate does he approach his image?
Sol: 
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE

Ques 3: A pole of height 4 m is kept in front of a vertical plane mirror of length 2 m. The lower end of the mirror is at a height of 6 m from the ground. The horizontal distance between the mirror and the pole is 2 m. Up to what minimum and maximum heights a man can see the image of top of the pole at a horizontal distance of 4 m (from the mirror) standing on the same horizontal line which is passing through the pole and the horizontal point below the mirror?
Sol:

DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
PQ = Pole
MN = Image of pole
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Minimum height required = AD = BD + AD = 10m
Further,
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Maximum height required = AE = BE+ AB = 16 m

Introductory Exercise 27.3

Ques 1: Assume that a certain spherical mirror has a focal length of- 10.0 cm. Locate and describe the image for object distances of
(a) 25.0 cm (b) 10.0 cm (c) 5.0 cm.
Sol:
Apply DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
(a) f = -10 cm, u = - 25 cm Solving we get, v = -16.7cm
Since, v is negative, image is in front of mirror. So it is real. Similarly, we can solve for other parts.

Ques 2: A ball is dropped from rest 3.0 m directly above the vertex of a concave mirror that has a radius of 1.0 m and lies in a horizontal plane.
(a) Describe the motion of ball's image in the mirror.
(b) At what time do the ball and its image coincide?
Sol: 
(a) From O to F, image is real.
From F to P, image is virtual.
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
At O
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
v = - 0.6 m

(b) When object is at C and P, image coincides with object. Using,
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
or  DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE

Ques 3: A spherical mirror is to be used to form on a screen 5.0 m from the object an image five times the size of the object.
(a) Describe the type of mirror required.
(b) Where should the mirror be positioned relative to the object?
Sol: 
(a) Image has to be taken on a screen. So it should be real. Hence mirror should be concave.
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Image is 5 times magnified. Hence, | v | = 5 | u |
or
(5 + x) = 5x
Solving we get, x = 1.25 cm
Now using, DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
We have, DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Solving we get, R = - 2.08 m

Ques 4: Figure shows two rays P and Q being reflected by a mirror and going as F and Q'. State which type of mirror is this?
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Sol:
Reflected rays are neither converging nor diverging. Hence, mirror is plane.

Ques 5: The following table shows object distance, object size, and mirror focal length (all in centimeters) for various situations. Use ray tracing to determine the image size and location for each case.
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Sol: 
Let us draw the ray diagram for the first part.
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
In ΔO1O2P and ΔP I1I2 we have,
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
or
x = (20)(I1,I2) ...(i)
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
In ΔO1O2C and l1I2 C we have, DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
or (40 - x) = 40(I1I2) ....(ii)
Solving Eqs. (i) and (ii) we get,  DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE

Ques 6: Complete the following table for spherical mirrors. All distances are in centrimeters.

Mirror Type

Radius

Focal Length

Object Distance

Image Distance

Lateral Magnification

Real/Virtual Image


-2540





20-30



+15
+30




20
-2.0

+20

-40



-10

+2.0

Sol: Negative focal length means concave mirror and positive focal length means convex mirror.
Apply, DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
And   DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE
Further, if v is negative then image is in front of mirror. So, it is real and inverted.
R = 2f

The document DC Pandey Solutions: Reflection of Light Notes | Study DC Pandey Solutions for NEET Physics - JEE is a part of the JEE Course DC Pandey Solutions for NEET Physics.
All you need of JEE at this link: JEE

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