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Ray Optics

282041 A linear object of height $10 cm$ is kept in front of a concave mirror of radius of curature 15 $cm$ , at a distance of $10 cm$ . The image formed is:

1 magnified and erect

2 magnified and inverted

3 diminished and erect

4 diminished and inverted

Explanation:

B: Given, Object height $(h_{o}) = 10 cm$
Radius of curvature of mirror $(R) = - 15 cm$
Focus of mirror $(f) = - \frac{15}{2} cm$
Object distance from the mirror $(u) = - 10 cm$
By using mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{\frac{15}{2}} = - \frac{1}{10} + \frac{1}{V} \\ \frac{1}{V} = \frac{1}{10} - \frac{2}{15} \\ \frac{1}{V} = \frac{3 - 4}{30} \\ v = - 30 cm \end{array}$
We know that,
$\begin{array}{r} Magnification (m) = - \frac{v}{u} = \frac{h_{i}}{h_{o}} \\ ∴ - \frac{(- 30)}{(- 10)} = \frac{h_{i}}{10} \\ h_{i} = - 30 cm \end{array}$
Hence, images are magnified and inverted.

Karnataka CET-2017

Ray Optics

282042 If an object is placed $10 cm$ in front of a concave mirror of focal length $20 cm$ , the image will be

1 diminished, upright, virtual

2 enlarged, upright, virtual

3 diminished, inverted, real

4 enlarged, upright, real

Explanation:

B: Given, object distance $= 10 cm$
Focal length $= 20 cm$
From the above value, it is clear that, object is placed between focal length and pole.
Hence, image is enlarge, upright and virtual
![original image](https://cdn.mathpix.com/snip/images/nckGOh6VZgAORaaNNegSw3wbP60g_qLO1lmpeUGX-uQ.original.fullsize.png)

CG PET- 2007

Ray Optics

282043 A point object is placed at a distance of $10 cm$ and its real image is formed at a distance of $20$ $cm$ from a concave mirror. If the object is moved by $0.1 cm$ towards the mirror, the image will shift by about

1

0.4 cm

away from the mirror

2

0.4 cm

towards the mirror

3

0.8 cm

away from the mirror

4

0.8 cm

towards the mirror

Explanation:

A: Given, object placed (u) $= - 10 cm$
Image formed at distance $(v) = - 20 cm$ (real)
For focal length of concave mirror,
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{10} - \frac{1}{20} \\ \frac{1}{f} = \frac{- 1 - 2}{20} = \frac{- 3}{20} \\ f = - \frac{20}{3} \end{array}$
When object move $0.1 cm$ toward the mirror then new position of object -
$u = 10 - 0.1 = 9.9 cm$
Then, new position of image -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u^{'}} + \frac{1}{V^{'}} \\ \frac{1}{- (\frac{20}{3})} = \frac{1}{- 9.9} + \frac{1}{V^{'}} \\ \frac{1}{V^{'}} = \frac{- 3}{20} + \frac{1}{9.9} \\ \frac{1}{V^{'}} = \frac{- 29.7 + 20}{20 \times 9.9} \end{array}$
$\begin{aligned} v^{'} & = \frac{20 \times 9.9}{- 9.7} \\ v^{'} & = \frac{- 198}{9.7} \\ v^{'} & = - 20.4 cm \end{aligned}$
Image shifted $= 20.4 - 20 = 0.4 cm$
Hence, image shifted $0.4 cm$ away from the mirror.

CG PET- 2005

Ray Optics

282044 Radius of curvature of concave mirror is $40 cm$ and the size of image of real is twice as that of object, then the object distance is

1

60 cm

2

20 cm

3

40 cm

4

30 cm

Explanation:

D: Given, radius of curvature (R) $= - 40 cm$
Then, focal length $= - \frac{40}{2} = - 20 cm$
Magnification $= - 2$ (for real image)
$\begin{array}{r} - 2 = \frac{- v}{u} \\ v = 2 u \end{array}$
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{20} = \frac{1}{u} + \frac{1}{2 u} \\ - \frac{1}{20} = \frac{3}{2 u} \\ u = - 30 cm \end{array}$
Hence, distance of object is $30 cm$ .

CG PET- 2007

Ray Optics

282045 A point object is moving on the principal axis of a concave mirror of focal length $24 cm$ towards the mirror. When it is at a distance of $60 cm$ from the mirror, its velocity is $9 cm / s$ . What is the velocity of the image at that instant?

1

5 cm / s

towards the mirror

2

4 cm / s

towards the mirror

3

4 cm / s

away from the mirror

4

9 cm / s

away from the mirror

Explanation:

C: Given, focal length (f) $= - 24 cm$
Object distance $(u) = - 60 cm$
And, initial velocity of object $(u_{i}) = 9 cm / s$
Final velocity of object $(v_{f}) =$ ?
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{(- 24)} = \frac{1}{(- 60)} + \frac{1}{v} \\ \frac{1}{v} = \frac{1}{60} - \frac{1}{24} \\ \frac{1}{V} = \frac{2 - 5}{120} \\ v = - \frac{120}{3} \\ v = - 40 cm \end{array}$
And, $\frac{1}{f} = \frac{1}{u} + \frac{1}{V}$
Now, differentiating the above equation with respect to time -
$\begin{array}{r} 0 = - \frac{1}{u^{2}} \frac{du}{dt} - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{1}{u^{2}} \frac{du}{dt} = - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{dv}{dt} = - \frac{v^{2}}{u^{2}} \frac{du}{dt} (∵ \frac{dv}{dt} = v_{f}) \\ v_{f} = - {(\frac{- 40}{- 60})}^{2} \times u_{i} \\ v_{f} = - \frac{4}{9} \times 9 \\ v_{f} = - 4 cm / s \end{array}$
Hence, velocity of image is $4 cm / s$ away from the mirror.

CG PET- 2004

Ray Optics

282041 A linear object of height $10 cm$ is kept in front of a concave mirror of radius of curature 15 $cm$ , at a distance of $10 cm$ . The image formed is:

1 magnified and erect

2 magnified and inverted

3 diminished and erect

4 diminished and inverted

Explanation:

B: Given, Object height $(h_{o}) = 10 cm$
Radius of curvature of mirror $(R) = - 15 cm$
Focus of mirror $(f) = - \frac{15}{2} cm$
Object distance from the mirror $(u) = - 10 cm$
By using mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{\frac{15}{2}} = - \frac{1}{10} + \frac{1}{V} \\ \frac{1}{V} = \frac{1}{10} - \frac{2}{15} \\ \frac{1}{V} = \frac{3 - 4}{30} \\ v = - 30 cm \end{array}$
We know that,
$\begin{array}{r} Magnification (m) = - \frac{v}{u} = \frac{h_{i}}{h_{o}} \\ ∴ - \frac{(- 30)}{(- 10)} = \frac{h_{i}}{10} \\ h_{i} = - 30 cm \end{array}$
Hence, images are magnified and inverted.

Karnataka CET-2017

Ray Optics

282042 If an object is placed $10 cm$ in front of a concave mirror of focal length $20 cm$ , the image will be

1 diminished, upright, virtual

2 enlarged, upright, virtual

3 diminished, inverted, real

4 enlarged, upright, real

Explanation:

B: Given, object distance $= 10 cm$
Focal length $= 20 cm$
From the above value, it is clear that, object is placed between focal length and pole.
Hence, image is enlarge, upright and virtual
![original image](https://cdn.mathpix.com/snip/images/nckGOh6VZgAORaaNNegSw3wbP60g_qLO1lmpeUGX-uQ.original.fullsize.png)

CG PET- 2007

Ray Optics

282043 A point object is placed at a distance of $10 cm$ and its real image is formed at a distance of $20$ $cm$ from a concave mirror. If the object is moved by $0.1 cm$ towards the mirror, the image will shift by about

1

0.4 cm

away from the mirror

2

0.4 cm

towards the mirror

3

0.8 cm

away from the mirror

4

0.8 cm

towards the mirror

Explanation:

A: Given, object placed (u) $= - 10 cm$
Image formed at distance $(v) = - 20 cm$ (real)
For focal length of concave mirror,
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{10} - \frac{1}{20} \\ \frac{1}{f} = \frac{- 1 - 2}{20} = \frac{- 3}{20} \\ f = - \frac{20}{3} \end{array}$
When object move $0.1 cm$ toward the mirror then new position of object -
$u = 10 - 0.1 = 9.9 cm$
Then, new position of image -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u^{'}} + \frac{1}{V^{'}} \\ \frac{1}{- (\frac{20}{3})} = \frac{1}{- 9.9} + \frac{1}{V^{'}} \\ \frac{1}{V^{'}} = \frac{- 3}{20} + \frac{1}{9.9} \\ \frac{1}{V^{'}} = \frac{- 29.7 + 20}{20 \times 9.9} \end{array}$
$\begin{aligned} v^{'} & = \frac{20 \times 9.9}{- 9.7} \\ v^{'} & = \frac{- 198}{9.7} \\ v^{'} & = - 20.4 cm \end{aligned}$
Image shifted $= 20.4 - 20 = 0.4 cm$
Hence, image shifted $0.4 cm$ away from the mirror.

CG PET- 2005

Ray Optics

282044 Radius of curvature of concave mirror is $40 cm$ and the size of image of real is twice as that of object, then the object distance is

1

60 cm

2

20 cm

3

40 cm

4

30 cm

Explanation:

D: Given, radius of curvature (R) $= - 40 cm$
Then, focal length $= - \frac{40}{2} = - 20 cm$
Magnification $= - 2$ (for real image)
$\begin{array}{r} - 2 = \frac{- v}{u} \\ v = 2 u \end{array}$
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{20} = \frac{1}{u} + \frac{1}{2 u} \\ - \frac{1}{20} = \frac{3}{2 u} \\ u = - 30 cm \end{array}$
Hence, distance of object is $30 cm$ .

CG PET- 2007

Ray Optics

282045 A point object is moving on the principal axis of a concave mirror of focal length $24 cm$ towards the mirror. When it is at a distance of $60 cm$ from the mirror, its velocity is $9 cm / s$ . What is the velocity of the image at that instant?

1

5 cm / s

towards the mirror

2

4 cm / s

towards the mirror

3

4 cm / s

away from the mirror

4

9 cm / s

away from the mirror

Explanation:

C: Given, focal length (f) $= - 24 cm$
Object distance $(u) = - 60 cm$
And, initial velocity of object $(u_{i}) = 9 cm / s$
Final velocity of object $(v_{f}) =$ ?
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{(- 24)} = \frac{1}{(- 60)} + \frac{1}{v} \\ \frac{1}{v} = \frac{1}{60} - \frac{1}{24} \\ \frac{1}{V} = \frac{2 - 5}{120} \\ v = - \frac{120}{3} \\ v = - 40 cm \end{array}$
And, $\frac{1}{f} = \frac{1}{u} + \frac{1}{V}$
Now, differentiating the above equation with respect to time -
$\begin{array}{r} 0 = - \frac{1}{u^{2}} \frac{du}{dt} - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{1}{u^{2}} \frac{du}{dt} = - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{dv}{dt} = - \frac{v^{2}}{u^{2}} \frac{du}{dt} (∵ \frac{dv}{dt} = v_{f}) \\ v_{f} = - {(\frac{- 40}{- 60})}^{2} \times u_{i} \\ v_{f} = - \frac{4}{9} \times 9 \\ v_{f} = - 4 cm / s \end{array}$
Hence, velocity of image is $4 cm / s$ away from the mirror.

CG PET- 2004

Ray Optics

282041 A linear object of height $10 cm$ is kept in front of a concave mirror of radius of curature 15 $cm$ , at a distance of $10 cm$ . The image formed is:

1 magnified and erect

2 magnified and inverted

3 diminished and erect

4 diminished and inverted

Explanation:

B: Given, Object height $(h_{o}) = 10 cm$
Radius of curvature of mirror $(R) = - 15 cm$
Focus of mirror $(f) = - \frac{15}{2} cm$
Object distance from the mirror $(u) = - 10 cm$
By using mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{\frac{15}{2}} = - \frac{1}{10} + \frac{1}{V} \\ \frac{1}{V} = \frac{1}{10} - \frac{2}{15} \\ \frac{1}{V} = \frac{3 - 4}{30} \\ v = - 30 cm \end{array}$
We know that,
$\begin{array}{r} Magnification (m) = - \frac{v}{u} = \frac{h_{i}}{h_{o}} \\ ∴ - \frac{(- 30)}{(- 10)} = \frac{h_{i}}{10} \\ h_{i} = - 30 cm \end{array}$
Hence, images are magnified and inverted.

Karnataka CET-2017

Ray Optics

282042 If an object is placed $10 cm$ in front of a concave mirror of focal length $20 cm$ , the image will be

1 diminished, upright, virtual

2 enlarged, upright, virtual

3 diminished, inverted, real

4 enlarged, upright, real

Explanation:

B: Given, object distance $= 10 cm$
Focal length $= 20 cm$
From the above value, it is clear that, object is placed between focal length and pole.
Hence, image is enlarge, upright and virtual
![original image](https://cdn.mathpix.com/snip/images/nckGOh6VZgAORaaNNegSw3wbP60g_qLO1lmpeUGX-uQ.original.fullsize.png)

CG PET- 2007

Ray Optics

282043 A point object is placed at a distance of $10 cm$ and its real image is formed at a distance of $20$ $cm$ from a concave mirror. If the object is moved by $0.1 cm$ towards the mirror, the image will shift by about

1

0.4 cm

away from the mirror

2

0.4 cm

towards the mirror

3

0.8 cm

away from the mirror

4

0.8 cm

towards the mirror

Explanation:

A: Given, object placed (u) $= - 10 cm$
Image formed at distance $(v) = - 20 cm$ (real)
For focal length of concave mirror,
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{10} - \frac{1}{20} \\ \frac{1}{f} = \frac{- 1 - 2}{20} = \frac{- 3}{20} \\ f = - \frac{20}{3} \end{array}$
When object move $0.1 cm$ toward the mirror then new position of object -
$u = 10 - 0.1 = 9.9 cm$
Then, new position of image -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u^{'}} + \frac{1}{V^{'}} \\ \frac{1}{- (\frac{20}{3})} = \frac{1}{- 9.9} + \frac{1}{V^{'}} \\ \frac{1}{V^{'}} = \frac{- 3}{20} + \frac{1}{9.9} \\ \frac{1}{V^{'}} = \frac{- 29.7 + 20}{20 \times 9.9} \end{array}$
$\begin{aligned} v^{'} & = \frac{20 \times 9.9}{- 9.7} \\ v^{'} & = \frac{- 198}{9.7} \\ v^{'} & = - 20.4 cm \end{aligned}$
Image shifted $= 20.4 - 20 = 0.4 cm$
Hence, image shifted $0.4 cm$ away from the mirror.

CG PET- 2005

Ray Optics

282044 Radius of curvature of concave mirror is $40 cm$ and the size of image of real is twice as that of object, then the object distance is

1

60 cm

2

20 cm

3

40 cm

4

30 cm

Explanation:

D: Given, radius of curvature (R) $= - 40 cm$
Then, focal length $= - \frac{40}{2} = - 20 cm$
Magnification $= - 2$ (for real image)
$\begin{array}{r} - 2 = \frac{- v}{u} \\ v = 2 u \end{array}$
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{20} = \frac{1}{u} + \frac{1}{2 u} \\ - \frac{1}{20} = \frac{3}{2 u} \\ u = - 30 cm \end{array}$
Hence, distance of object is $30 cm$ .

CG PET- 2007

Ray Optics

282045 A point object is moving on the principal axis of a concave mirror of focal length $24 cm$ towards the mirror. When it is at a distance of $60 cm$ from the mirror, its velocity is $9 cm / s$ . What is the velocity of the image at that instant?

1

5 cm / s

towards the mirror

2

4 cm / s

towards the mirror

3

4 cm / s

away from the mirror

4

9 cm / s

away from the mirror

Explanation:

C: Given, focal length (f) $= - 24 cm$
Object distance $(u) = - 60 cm$
And, initial velocity of object $(u_{i}) = 9 cm / s$
Final velocity of object $(v_{f}) =$ ?
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{(- 24)} = \frac{1}{(- 60)} + \frac{1}{v} \\ \frac{1}{v} = \frac{1}{60} - \frac{1}{24} \\ \frac{1}{V} = \frac{2 - 5}{120} \\ v = - \frac{120}{3} \\ v = - 40 cm \end{array}$
And, $\frac{1}{f} = \frac{1}{u} + \frac{1}{V}$
Now, differentiating the above equation with respect to time -
$\begin{array}{r} 0 = - \frac{1}{u^{2}} \frac{du}{dt} - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{1}{u^{2}} \frac{du}{dt} = - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{dv}{dt} = - \frac{v^{2}}{u^{2}} \frac{du}{dt} (∵ \frac{dv}{dt} = v_{f}) \\ v_{f} = - {(\frac{- 40}{- 60})}^{2} \times u_{i} \\ v_{f} = - \frac{4}{9} \times 9 \\ v_{f} = - 4 cm / s \end{array}$
Hence, velocity of image is $4 cm / s$ away from the mirror.

CG PET- 2004

Ray Optics

282041 A linear object of height $10 cm$ is kept in front of a concave mirror of radius of curature 15 $cm$ , at a distance of $10 cm$ . The image formed is:

1 magnified and erect

2 magnified and inverted

3 diminished and erect

4 diminished and inverted

Explanation:

B: Given, Object height $(h_{o}) = 10 cm$
Radius of curvature of mirror $(R) = - 15 cm$
Focus of mirror $(f) = - \frac{15}{2} cm$
Object distance from the mirror $(u) = - 10 cm$
By using mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{\frac{15}{2}} = - \frac{1}{10} + \frac{1}{V} \\ \frac{1}{V} = \frac{1}{10} - \frac{2}{15} \\ \frac{1}{V} = \frac{3 - 4}{30} \\ v = - 30 cm \end{array}$
We know that,
$\begin{array}{r} Magnification (m) = - \frac{v}{u} = \frac{h_{i}}{h_{o}} \\ ∴ - \frac{(- 30)}{(- 10)} = \frac{h_{i}}{10} \\ h_{i} = - 30 cm \end{array}$
Hence, images are magnified and inverted.

Karnataka CET-2017

Ray Optics

282042 If an object is placed $10 cm$ in front of a concave mirror of focal length $20 cm$ , the image will be

1 diminished, upright, virtual

2 enlarged, upright, virtual

3 diminished, inverted, real

4 enlarged, upright, real

Explanation:

B: Given, object distance $= 10 cm$
Focal length $= 20 cm$
From the above value, it is clear that, object is placed between focal length and pole.
Hence, image is enlarge, upright and virtual
![original image](https://cdn.mathpix.com/snip/images/nckGOh6VZgAORaaNNegSw3wbP60g_qLO1lmpeUGX-uQ.original.fullsize.png)

CG PET- 2007

Ray Optics

282043 A point object is placed at a distance of $10 cm$ and its real image is formed at a distance of $20$ $cm$ from a concave mirror. If the object is moved by $0.1 cm$ towards the mirror, the image will shift by about

1

0.4 cm

away from the mirror

2

0.4 cm

towards the mirror

3

0.8 cm

away from the mirror

4

0.8 cm

towards the mirror

Explanation:

A: Given, object placed (u) $= - 10 cm$
Image formed at distance $(v) = - 20 cm$ (real)
For focal length of concave mirror,
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{10} - \frac{1}{20} \\ \frac{1}{f} = \frac{- 1 - 2}{20} = \frac{- 3}{20} \\ f = - \frac{20}{3} \end{array}$
When object move $0.1 cm$ toward the mirror then new position of object -
$u = 10 - 0.1 = 9.9 cm$
Then, new position of image -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u^{'}} + \frac{1}{V^{'}} \\ \frac{1}{- (\frac{20}{3})} = \frac{1}{- 9.9} + \frac{1}{V^{'}} \\ \frac{1}{V^{'}} = \frac{- 3}{20} + \frac{1}{9.9} \\ \frac{1}{V^{'}} = \frac{- 29.7 + 20}{20 \times 9.9} \end{array}$
$\begin{aligned} v^{'} & = \frac{20 \times 9.9}{- 9.7} \\ v^{'} & = \frac{- 198}{9.7} \\ v^{'} & = - 20.4 cm \end{aligned}$
Image shifted $= 20.4 - 20 = 0.4 cm$
Hence, image shifted $0.4 cm$ away from the mirror.

CG PET- 2005

Ray Optics

282044 Radius of curvature of concave mirror is $40 cm$ and the size of image of real is twice as that of object, then the object distance is

1

60 cm

2

20 cm

3

40 cm

4

30 cm

Explanation:

D: Given, radius of curvature (R) $= - 40 cm$
Then, focal length $= - \frac{40}{2} = - 20 cm$
Magnification $= - 2$ (for real image)
$\begin{array}{r} - 2 = \frac{- v}{u} \\ v = 2 u \end{array}$
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{20} = \frac{1}{u} + \frac{1}{2 u} \\ - \frac{1}{20} = \frac{3}{2 u} \\ u = - 30 cm \end{array}$
Hence, distance of object is $30 cm$ .

CG PET- 2007

Ray Optics

282045 A point object is moving on the principal axis of a concave mirror of focal length $24 cm$ towards the mirror. When it is at a distance of $60 cm$ from the mirror, its velocity is $9 cm / s$ . What is the velocity of the image at that instant?

1

5 cm / s

towards the mirror

2

4 cm / s

towards the mirror

3

4 cm / s

away from the mirror

4

9 cm / s

away from the mirror

Explanation:

C: Given, focal length (f) $= - 24 cm$
Object distance $(u) = - 60 cm$
And, initial velocity of object $(u_{i}) = 9 cm / s$
Final velocity of object $(v_{f}) =$ ?
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{(- 24)} = \frac{1}{(- 60)} + \frac{1}{v} \\ \frac{1}{v} = \frac{1}{60} - \frac{1}{24} \\ \frac{1}{V} = \frac{2 - 5}{120} \\ v = - \frac{120}{3} \\ v = - 40 cm \end{array}$
And, $\frac{1}{f} = \frac{1}{u} + \frac{1}{V}$
Now, differentiating the above equation with respect to time -
$\begin{array}{r} 0 = - \frac{1}{u^{2}} \frac{du}{dt} - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{1}{u^{2}} \frac{du}{dt} = - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{dv}{dt} = - \frac{v^{2}}{u^{2}} \frac{du}{dt} (∵ \frac{dv}{dt} = v_{f}) \\ v_{f} = - {(\frac{- 40}{- 60})}^{2} \times u_{i} \\ v_{f} = - \frac{4}{9} \times 9 \\ v_{f} = - 4 cm / s \end{array}$
Hence, velocity of image is $4 cm / s$ away from the mirror.

CG PET- 2004

Ray Optics

282041 A linear object of height $10 cm$ is kept in front of a concave mirror of radius of curature 15 $cm$ , at a distance of $10 cm$ . The image formed is:

1 magnified and erect

2 magnified and inverted

3 diminished and erect

4 diminished and inverted

Explanation:

B: Given, Object height $(h_{o}) = 10 cm$
Radius of curvature of mirror $(R) = - 15 cm$
Focus of mirror $(f) = - \frac{15}{2} cm$
Object distance from the mirror $(u) = - 10 cm$
By using mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{\frac{15}{2}} = - \frac{1}{10} + \frac{1}{V} \\ \frac{1}{V} = \frac{1}{10} - \frac{2}{15} \\ \frac{1}{V} = \frac{3 - 4}{30} \\ v = - 30 cm \end{array}$
We know that,
$\begin{array}{r} Magnification (m) = - \frac{v}{u} = \frac{h_{i}}{h_{o}} \\ ∴ - \frac{(- 30)}{(- 10)} = \frac{h_{i}}{10} \\ h_{i} = - 30 cm \end{array}$
Hence, images are magnified and inverted.

Karnataka CET-2017

Ray Optics

282042 If an object is placed $10 cm$ in front of a concave mirror of focal length $20 cm$ , the image will be

1 diminished, upright, virtual

2 enlarged, upright, virtual

3 diminished, inverted, real

4 enlarged, upright, real

Explanation:

B: Given, object distance $= 10 cm$
Focal length $= 20 cm$
From the above value, it is clear that, object is placed between focal length and pole.
Hence, image is enlarge, upright and virtual
![original image](https://cdn.mathpix.com/snip/images/nckGOh6VZgAORaaNNegSw3wbP60g_qLO1lmpeUGX-uQ.original.fullsize.png)

CG PET- 2007

Ray Optics

282043 A point object is placed at a distance of $10 cm$ and its real image is formed at a distance of $20$ $cm$ from a concave mirror. If the object is moved by $0.1 cm$ towards the mirror, the image will shift by about

1

0.4 cm

away from the mirror

2

0.4 cm

towards the mirror

3

0.8 cm

away from the mirror

4

0.8 cm

towards the mirror

Explanation:

A: Given, object placed (u) $= - 10 cm$
Image formed at distance $(v) = - 20 cm$ (real)
For focal length of concave mirror,
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{f} = - \frac{1}{10} - \frac{1}{20} \\ \frac{1}{f} = \frac{- 1 - 2}{20} = \frac{- 3}{20} \\ f = - \frac{20}{3} \end{array}$
When object move $0.1 cm$ toward the mirror then new position of object -
$u = 10 - 0.1 = 9.9 cm$
Then, new position of image -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u^{'}} + \frac{1}{V^{'}} \\ \frac{1}{- (\frac{20}{3})} = \frac{1}{- 9.9} + \frac{1}{V^{'}} \\ \frac{1}{V^{'}} = \frac{- 3}{20} + \frac{1}{9.9} \\ \frac{1}{V^{'}} = \frac{- 29.7 + 20}{20 \times 9.9} \end{array}$
$\begin{aligned} v^{'} & = \frac{20 \times 9.9}{- 9.7} \\ v^{'} & = \frac{- 198}{9.7} \\ v^{'} & = - 20.4 cm \end{aligned}$
Image shifted $= 20.4 - 20 = 0.4 cm$
Hence, image shifted $0.4 cm$ away from the mirror.

CG PET- 2005

Ray Optics

282044 Radius of curvature of concave mirror is $40 cm$ and the size of image of real is twice as that of object, then the object distance is

1

60 cm

2

20 cm

3

40 cm

4

30 cm

Explanation:

D: Given, radius of curvature (R) $= - 40 cm$
Then, focal length $= - \frac{40}{2} = - 20 cm$
Magnification $= - 2$ (for real image)
$\begin{array}{r} - 2 = \frac{- v}{u} \\ v = 2 u \end{array}$
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ - \frac{1}{20} = \frac{1}{u} + \frac{1}{2 u} \\ - \frac{1}{20} = \frac{3}{2 u} \\ u = - 30 cm \end{array}$
Hence, distance of object is $30 cm$ .

CG PET- 2007

Ray Optics

282045 A point object is moving on the principal axis of a concave mirror of focal length $24 cm$ towards the mirror. When it is at a distance of $60 cm$ from the mirror, its velocity is $9 cm / s$ . What is the velocity of the image at that instant?

1

5 cm / s

towards the mirror

2

4 cm / s

towards the mirror

3

4 cm / s

away from the mirror

4

9 cm / s

away from the mirror

Explanation:

C: Given, focal length (f) $= - 24 cm$
Object distance $(u) = - 60 cm$
And, initial velocity of object $(u_{i}) = 9 cm / s$
Final velocity of object $(v_{f}) =$ ?
From mirror formula -
$\begin{array}{r} \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ \frac{1}{(- 24)} = \frac{1}{(- 60)} + \frac{1}{v} \\ \frac{1}{v} = \frac{1}{60} - \frac{1}{24} \\ \frac{1}{V} = \frac{2 - 5}{120} \\ v = - \frac{120}{3} \\ v = - 40 cm \end{array}$
And, $\frac{1}{f} = \frac{1}{u} + \frac{1}{V}$
Now, differentiating the above equation with respect to time -
$\begin{array}{r} 0 = - \frac{1}{u^{2}} \frac{du}{dt} - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{1}{u^{2}} \frac{du}{dt} = - \frac{1}{v^{2}} \frac{dv}{dt} \\ \frac{dv}{dt} = - \frac{v^{2}}{u^{2}} \frac{du}{dt} (∵ \frac{dv}{dt} = v_{f}) \\ v_{f} = - {(\frac{- 40}{- 60})}^{2} \times u_{i} \\ v_{f} = - \frac{4}{9} \times 9 \\ v_{f} = - 4 cm / s \end{array}$
Hence, velocity of image is $4 cm / s$ away from the mirror.

CG PET- 2004