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Physics Part-II

Ray Optics and Optical Instruments

Use the mirror equation to deduce that: (a) an object placed b

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

(a) A giant refracting telescope at an observatory has an objective lens of focal length

15 m

. If an eyepiece of focal length

1.0

cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is

3.48 \times 1 0^{6} m

, and the radius of lunar orbit is

3.8 \times 1 0^{8} m

Use the mirror equation to deduce that:
(a) an object placed between

f

and

2 f

of a concave mirror produces a real image beyond

2 f

.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass

=

1.5. Does the answer depend on the location of the slab?

View all question

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Chapter 6: Semiconductor Electronics: Materials, Devices and Simple Circuits

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Question

Easy

Solving time: 6 mins

Use the mirror equation to deduce that:
(a) an object placed between $f$ and $2 f$ of a concave mirror produces a real image beyond $2 f$ .
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Views: 5,750 students

Updated on: Jan 23, 2024

Book:

Physics Part-II (NCERT)

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Use the mirror equation to deduce that:
(a) an object placed between

f

and

2 f

of a concave mirror produces a real image beyond

2 f

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

a) For a concave mirror, the focal length (

f

) is negative.

When the object is placed on the left side of the mirror, the object distance (

u

) is negative

For image distance

v

we can write

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

...(1)

The object lies between

f

and

2 f

2 f < u < f

\frac{1}{2 f} > \frac{1}{u} > \frac{1}{f}

\frac{- 1}{2 f} < \frac{- 1}{u} < \frac{- 1}{f}

\frac{1}{f} - \frac{1}{2 f} < \frac{1}{f} - \frac{1}{u}

.......

(2)

from eq. 1

\frac{1}{2 f} < \frac{1}{v} < 0

;

v

is negative

2 f > v

- v > - 2 f

Therefore, the image lies beyond

2 f

b) For a convex mirror, the focal length (

f

) is positive

When the object is placed on the left side of the mirror, the object distance (

u

) is negative

For image distance

v

, we have the mirror formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Using eq (

2

)

\frac{1}{v} < 0

v > 0

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

c) For a convex mirror, the focal length (f) is positive

When the object is placed on the left side of the mirror, the object distance (u) is negative

For image distance

v

, we have the mirror formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

but

u < 0

therefore

\frac{1}{v} > \frac{1}{f}

v < f

Hence, the image formed is diminished and is located between the focus (

f

) and the pole

d) For a concave mirror, the focal length (

f

) is negative

When the object is placed on the left side of the mirror, the object distance (

u

) is negative

It is placed between the focus (f) and the pole

therefore

f > u > 0

\frac{1}{f} < \frac{1}{u} < 0

\frac{1}{f} - \frac{1}{u}

For image distance

v

, we have the mirror formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

\frac{1}{v} < 0

v > 0

The image is formed on the right side of the mirror. Hence, it is a virtual image

For

u < 0

and

v > 0

, we can write:

\frac{1}{u} > \frac{1}{v}

v > u

magnification

m = \frac{v}{u} > 1

Hence, the formed image is enlarged.

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Practice more questions from Physics Part-II (NCERT)

(a) A giant refracting telescope at an observatory has an objective lens of focal length

15 m

. If an eyepiece of focal length

1.0

3.48 \times 1 0^{6} m

, and the radius of lunar orbit is

3.8 \times 1 0^{8} m

Use the mirror equation to deduce that:
(a) an object placed between

f

and

2 f

of a concave mirror produces a real image beyond

2 f

=

1.5. Does the answer depend on the location of the slab?

View all

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Question Text	Use the mirror equation to deduce that: (a) an object placed between $f$ and $2 f$ of a concave mirror produces a real image beyond $2 f$ . (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Updated On	Jan 23, 2024
Topic	Ray Optics and Optical Instruments
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 14
Upvotes	1372
Avg. Video Duration	10 min

Text solutionVerified

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