Question

1. An object of size 9 cm is placed at 20 cm in front of a concave mirror of focal length
15 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed
image can be obtained? Find the size and the nature of the image​

Answer 1

Answer

❀ Ho = 3 cm

❀ U = -20 [ sign convention ]

❀ F = -15 [ sign convention]

᯾ V = ?? ᯾

Using mirror formula ,

✪ 1/f = 1/U + 1/V

1/v = 1/U-1/f => 1/v = 1/(-20)-1/(-15)

1/v = 1/60 => V= -60

We know that,

M = -v/u = Hi/Ho . [ Where Hi is height of image and Ho is height of object ]

-v/u = Hi/Ho

-(-60)/(-20)

Hi/3=> Hi = -9

M is negative, so image is inverted and real . ✔︎✔︎

______________________________________________

_______________________________________________

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Height of object, (h1) = 9 cm
• Focal length of a concave mirror, (f) = -15 cm
• Distance of object from mirror, (u) = -20 cm

$\underline{\bf{Explanation\::}}$

As we know that formula of the mirror;

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

A/q

$\mapsto\tt{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }$

$\mapsto\tt{\dfrac{1}{-15} = \dfrac{1}{v} + \dfrac{1}{(-20)} }$

$\mapsto\tt{\dfrac{1}{-15} = \dfrac{1}{v} - \dfrac{1}{20} }$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{1}{-15} + \dfrac{1}{20} }$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{-4 + 3}{60}}$

$\mapsto\tt{\dfrac{1}{v} = \dfrac{-1}{60}}$

$\mapsto\tt{-v = 60}$

$\mapsto\bf{v = -60\:cm}$

∴ The screen should be kept at a distance of 60 cm in front of mirror.

Now, as we know that formula of magnification.

$\bf{m = \dfrac{Height\:of\:Image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance \:of\:Image}{Distance\:of\:Object} =\dfrac{v}{u} }$

$\longrightarrow\tt{\dfrac{h_2}{h_1} =\dfrac{-v}{u} }$

$\longrightarrow\tt{\dfrac{h_2}{9} =\cancel{\dfrac{-60}{-20} }}$

$\longrightarrow\tt{\dfrac{h_2}{9} =3 }$

$\longrightarrow\tt{h_2 = 9 \times 3}$

$\longrightarrow\bf{h_2 = 27\:cm}$

Thus,

The height of Image will be 27 cm & the image is real and Inverted .