Question

an object 2cm is placed at a distance of 16cm from a concave mirror which produces a real image 3cm high then find the focal length and position of image​

Answer 1

Given that, an object 2cm is placed at a distance of 16cm from a concave mirror which produces a real image 3cm high.

{ ∴ u = -16 cm, ho = 2 cm and hi = -3 cm }

We have to find the focal length (f) and position of the image (v).

Now,

m = hi/ho = -v/u

Here; m = magnification, hi = height of image, ho = height of object, v = image distance from the mirror and u = object distance from the mirror.

Substitute the known values,

→ -3/2 = -v/(-16)

→ -3/2 = v/16

→ -3 = v/8

→ -3*8 = v

→ -24 = v

(Object and image are on the same side.)

Therefore, the image distance from the mirror is 24 cm.

Using Mirror Formula,

1/f = 1/v + 1/u

Substitute the known values,

→ 1/f = 1/(-24) + 1/(-16)

→ 1/f = -1/24 - 1/16

→ 1/f = (-2 - 3)/48

→ 1/f = -5/48

→ f = -48/5

→ f = -9.6

Therefore, the focal length of the concave mirror is -9.6 cm.

Answer 2

GiVeN :-

Object size ($\sf{h_o}$) = 2 cm

Object distance (u) = -16 cm

Image height ($\sf{h_i}$) = 3 cm

To DeTeRmInE :-

The image distance (v) and the focal length (f).

AcKnOwLeDgEmEnT :-

$\bigstar$ MIRROR FORMULA :

$\sf{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} }$

$\bigstar$ Magnification of a concave mirror :

$\sf{m=\dfrac{-v}{u}=\dfrac{h_i}{h_o} }$

SoLuTiOn :-

Substituting the values  :-

$\sf{m=\dfrac{-v}{u}=\dfrac{h_i}{h_o} }$

$\sf{\leadsto \dfrac{-v}{-16}=\dfrac{-3}{2} }\\\\\textsf{As the magnification is negative}\textsf{ for producing a real image.}$

$\sf{\leadsto \dfrac{v}{16} =\dfrac{-3}{2} }$

$\sf{\leadsto v=-3\times8 }\\\\\boxed{\sf{\leadsto v=-24\ cm}}$

$\rule{100}2$

Now, substituting the values in the MIRROR FORMULA :-

$\sf{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f} }$

$\sf{\leadsto \dfrac{1}{-24}+\dfrac{1}{-16}=\dfrac{1}{f} }\\\\\\\sf{\leadsto \dfrac{-1}{24}-\dfrac{1}{16}=\dfrac{1}{f} }\\\\\\\sf{\leadsto \dfrac{-2-3}{48}=\dfrac{1}{f} }\\\\\\\sf{\leadsto \dfrac{-5}{48}=\dfrac{1}{f} }\\\\\\\sf{\leadsto f=\dfrac{-48}{5} }\\\\\\\boxed{\sf{\leadsto f=-9.6\ cm}}$

Therefore, the position of the image is 24 cm in front of the mirror (left side) and the focal length of the mirror is -9.6 cm.