Question

An object 3cm high is placed at a distance of 8 cm from a concave mirror which produces a virtual image 5.5cm high. 1. what is the focal length of the mirror.
2. What is the position of image?​

Answer 1

Given :-

Mirror is concave

Height of the object (ho= 3 cm)

Object distance (u = 8 cm)

Height of image is (hi = 5.5 cm)

To find :-

Focal length of the mirror

Position of image.

Solution :-

According to sign convention,

u = -8 cm.

Firstly let's find the value of v i.e image distance by Magnification formula .

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

$\implies \: \sf \: \dfrac{ v}{8} = \dfrac{5.5}{3}$

$\implies \sf \: 3v = 5.5 \times 8$

$\implies \: \sf \: 3v = 44$

$\implies \: \sf \: v = \dfrac{44}{3}$

So, the image distance (v) = 44/3 cm = 14.6 cm

Now, finding the focal length of the mirror by using mirror formula.

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

$\implies \: \sf \dfrac{1}{f} = \dfrac{1}{ \dfrac{44}{3} } - \dfrac{1}{8}$

$\implies \: \sf \dfrac{1}{f} = \dfrac{3}{44} - \dfrac{1}{8}$

$\implies \: \sf \dfrac{1}{f} = \dfrac{6 - 11}{88}$

$\implies \: \sf \dfrac{1}{f} = \dfrac{ - 5}{88}$

$\sf \: f = \dfrac{ - 88}{5}$

$\sf \: f = \: - 17.6cm$

So, the focal length of mirror is -17.6 cm.

Position of image (v) = 14.6 cm . [Behind the mirror]

Answer 2

Answer:

1)f = -17.6cm

2)Behind the mirror

Explanation:

Given,

★Height of the object (h_o) = 3cm

★Height of the image(h_i) = 5.5cm

★ Object distance (u) = 8cm

Image formed is Virtual Image.

To Find :-

1)Focal length of the mirror

2)Position of the Image

Solution :-

Object distance(u) = 8cm

We know that 'u' will be negative in concave mirror

→ u = -8cm.

As we need to find focal length first we need to find both the image distance and the object distance. So let us find the image distance(v)

We know that,

Magnification of the mirrior = -v/u = h_i/h_o

→ -v/-8 = 5.5cm/3cm

v/8cm = 5.5/3

v = 5.5(8)/3

v = 44/3cm.

Now substituting the value of image distance in mirror formula :-

1/f = 1/u + 1/v

= 1/(-8)+1/44/3

= -1/8+3/44

= -1(11)+3(2)/88

= -11+6/88

1/f = -5/88 cm

→ f = -88/5cm

f = -17.6cm

We know that concave mirror forms virtual image only when the object is placed at between pole and focal length.

→ We can say that image is formed at behind the mirror

Therefore, Position of the mirrior is behind the mirror.

Formula Used :-

1) Magnification of the mirrior = -v/u = h_i/h_o

2) Mirror formula :-

1/f = 1/u + 1/v