Question

An object 2 cm is placed 30cm in front of a concave mirror of focal length 15cm.At what distance from the mirror should a screen be placed in order to obtain a sharp image? What will be the size of the image ​

Answer 1

According to the question:

Object distance, u=−30 cm

Focal length, f=−15 cm

Let the Image distance be v.

By mirror formula:

v

1

+

u

1

=

f

1

[4pt]

⇒

v

1

+

−30 cm

1

=

−15 cm

1

[4pt]

⇒

v

1

=−

−30 cm

1

+

−15 cm

1

[4pt]

⇒

v

1

=

30 cm

1−2

[4pt]

⇒

v

1

=−

30 cm

1

[4pt]

∴v=−30 cm

Thus, screen should be placed 30 cm in front of the mirror (Centre of curvature) to obtain the real image.

Now,

Height of object, h

1

=2 cm

Magnification, m=

h

1

h

2

=−

u

v

Putting values of v and u:

Magnification m=

2 cm

h

2

=−

−30 cm

−30 cm

⇒

2 cm

h

2

=−1

[4pt];

⇒h

2

=−1×2 cm=−2 cm

Thus, the height of the image is 2 cm and the negative sign means the image is inverted.

Thus real, inverted image of size same as that of object is formed.

Explanation:

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Answer 2

Given:-

$\sf \footnotesize{height \: of \: object , h_1 = 2cm}$

$\sf \footnotesize{Object \: distance, u= 30cm}$

$\sf \footnotesize{Focal \: length,f = 15cm}$

Required Answer:-

Distance at which the at which the screen should be placed to obtain a sharp image. In other words, the image distance,v.

Solution:-

$\sf \footnotesize{u = 30 cm}$

$\sf \footnotesize{f = 15cm}$

From mirror formula,

$\bf \footnotesize{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

$\sf \footnotesize \implies{ \frac{1}{15} = \frac{1}{30} + \frac{1}{v} }$

$\sf \footnotesize \implies{ \frac{1}{15} - \frac{1}{30} = \frac{1}{v} }$

$\sf \footnotesize{ \implies{ \frac{1}{v} = \frac{2 - 1}{30}}}$

$\sf \footnotesize{ \implies{ \frac{1}{v } = \frac{1}{30} }}$

$\sf \footnotesize{ \implies{v = 30 \: cm}}$

Hence, the screen should be placed at

30 cm from the mirror.