Question

1. An object 5 cm high forms a virtual image of 1.25 cm high, when placed in front of a convex
mirror at a distance of 24 cm. Calculate : (i) the position of the image, (ii) the focal length of
the convex mirror​

Answer 1

Given:-

→ Height of the object = 5 cm

→ Height of the image = 1.25 cm

→ Distance of the object = 24 cm

To find:-

→ Position of image.

→ Focal length of the convex mirror.

Solution:-

We know that :-

m = -(v/u) = hᵢ/hₒ

⇒ -(v/u) = hᵢ/hₒ

⇒ -v/(-24) = 1.25/5

⇒ -24(1.25) = 5(-v)

⇒ -30 = -5v

⇒ v = -30/-5

⇒ v = 6 cm

Hence, the image is formed at a distance of 6 cm from the mirror. The positive sign of v shows that the image is formed on the right i.e. behind the mirror.

________________________________

Since the mirror is convex, we have :-

• u = -24 cm

Now, according to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/6 + 1/(-24) = 1/f

⇒ 1/6 - 1/24 = 1/f

⇒ [4 - 1]/24 = 1/f

⇒ 3/24 = 1/f

⇒ 3f = 24

⇒ f = 24/3

⇒ f = 8 cm

Thus, focal length of the mirror is 8 cm .

Answer 2

Given :-

• As object 5 cm high forms a virtual image of 1.35 cm high , when placed in front of a convex mirror at a distance of 24 cm .

To Find :-

1. The position of the image ,
2. The focal length of the concex mirror.

Formula Used :-

$\clubsuit$ Magnification Formula :

$\longmapsto \sf\boxed{\bold{\pink{Magnification\: (m)\: =\: \dfrac{h_i}{h_o} =\: - \dfrac{v}{u}}}}$

where,

• $\sf h_i$ = Image Height
• $\sf h_o$ = Object Height
• v = Image Distance
• u = Object Distance

$\clubsuit$ Mirror Formula :

$\longmapsto \sf\boxed{\bold{\pink{\dfrac{1}{u} + \dfrac{1}{v} =\: \dfrac{1}{f}}}}$

where,

• u = Object Distance
• v = Image Distance
• f = Focal Length

Solution :-

i) First, we have to find the position of the image :

Given :

• Height of the image = 1.25 cm
• Height of the object = 5 cm
• Distance = 24 cm

According to the question by using the formula we,

$\implies \sf \dfrac{1.25}{5} =\: - \dfrac{v}{24}$

$\implies \sf \dfrac{1.25}{5} =\: \dfrac{\cancel{-} v}{\cancel{-} 24}$

$\implies \sf v =\: \dfrac{24 \times 1.25}{5}$

$\implies \sf v =\: \dfrac{24 \times 125}{5 \times 100}$

$\implies \sf v =\: \dfrac{30\cancel{00}}{5\cancel{00}}$

$\implies \sf v =\: \dfrac{\cancel{30}}{\cancel{5}}$

$\implies \sf \bold{\red{v =\: 6\: cm}}$

$\therefore$ The image is formed in 6 cm that is in front of the mirror.

$\rule{150}{2}$

ii) Now, we have to find the focal length of the convex mirror :

Given :

• Object Distance (u) = - 24
• Image Distance (v) = 6 cm

According to the question by using the formula we get,

$\implies \sf \dfrac{1}{- 24} + \dfrac{1}{6} =\: \dfrac{1}{f}$

$\implies \sf \dfrac{1}{f} =\: - \dfrac{1}{24} + \dfrac{1}{6}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{- 1 + 4}{24}$

$\implies \sf \dfrac{1}{f} =\: \dfrac{3}{24}$

By doing cross multiplication we get,

$\implies \sf 3f =\: 24$

$\implies \sf f =\: \dfrac{\cancel{24}}{\cancel{3}}$

$\implies \sf\bold{\red{f =\: 8\: cm}}$

$\therefore$ The focal length of the convex mirror is 8 cm .