Question

An object of size 7.0 cm is placed at a distance of 27 cm in front of concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and the nature of the image.

Answer 1

Answer:

Given:

size of object, o=+7cm

distance of object ,u=−27cm

focal length of concave mirror, f=−18cm

let us take size of image=I

so, mirror formula is

v

1

​  

+  

u

1

​  

=  

f

1

​  

 

so, putting values of u and v,

v

1

​  

=  

−18

1

​  

−  

−27

1

​  

 

v

1

​  

=  

−54

1

​  

 

v=−54cm

so, image is formed on object side only.

magnification,

m=  

u

−v

​  

=  

o

I

​  

 

m=−  

−27

−54

​  

=  

7

I

​  

 

I=−14cm

so, image is double in size to that of object.

Nature: real, inverted and magnified image.Given:

size of object, o=+7cm

distance of object ,u=−27cm

focal length of concave mirror, f=−18cm

let us take size of image=I

so, mirror formula is

v

1

​  

+  

u

1

​  

=  

f

1

​  

 

so, putting values of u and v,

v

1

​  

=  

−18

1

​  

−  

−27

1

​  

 

v

1

​  

=  

−54

1

​  

 

v=−54cm

so, image is formed on object side only.

magnification,

m=  

u

−v

​  

=  

o

I

​  

 

m=−  

−27

−54

​  

=  

7

I

​  

 

I=−14cm

so, image is double in size to that of object.

Nature: real, inverted and magnified image.

Explanation:

Answer 2

•GIVEN:-

$\bf\:➠ Object \: distance \: ,u=-27cm$

$\bf {➠Object \: height,h=7cm}$

$\bf ➠Focal \: length=-18cm$

$\bf \large\underline{\overline{FORMULA \: USED⇓⇓}}$

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

SOLUTION :-

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

$\frac{1}{v} = \bf \: \frac{1}{( - 18)} - \frac{1}{ (- 27)}$

$\frac{1}{v} = \bf \: \frac{1}{( - 18)} + \frac{1}{27}$

$\frac{1}{v} = \frac{( - 3 + 2)}{54} = \frac{ - 1 \: \: }{54 }$

$\boxed { \therefore \: v = \bf \: 54cm }$

•The screen should be placed at a distance of 54cm in front of the given mirror

$\bf \: ➽Magnification(m) = \frac{Image \: distance}{Object \: distance}$

$\bf \: m = \frac{ - 54}{ - 27} = - 2$

•Negative value of magnification shows that image is real.

$\bf \: m = \frac{Height \: of \: image}{Height \: of \: object} = \frac{h_1}{h}$

$\bf ➠h_1=m×h=(-2)×7=-14cm \\ \\ </p><p></p><p>$

➽The height of the image is negative which shows that image is inverted.

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