Question

an object is kept at a distance of 10cm from a convex mirror whose focal length is 15cm . find the position, nature and magnification​

Answer 1

Answer :-

(•) Image distance = 30 cm

(•) Nature is virtual and erect.

(•) magnification = 0.6

To Find :-

→ Position of image ,nature and magnification.

Explanation :-

Given that ,

• Focal length (f) = 15 cm

• object distance (u) = -10 cm ( left )

• image distance (v) = ?

• magnification (m) = ?

Apply mirror formula

$\implies \boxed{ \sf{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }} \\ \therefore \: \\ \\ \implies \sf{ \frac{1}{ 15} = \frac{1}{v} + \frac{1}{ - 10} } \\ \\ \implies \sf{\frac{1}{v} = \frac{1}{10} + \frac{1}{15} } \\ \\ \implies \sf{ \frac{1}{v} = \frac{3 + 2}{30} } \\ \\ \implies \sf{ \frac{1}{v} = \frac{5}{30} } \\ \\ \implies \boxed{ \sf{v(image \: distance) = 6 \: cm \: }}$

Nature :-

→ Because v is positive hence image is erect and virtual .

Magnification :-

We know that ;

$\to \sf{magnification \: = \frac{ - v}{u} = \frac{ - 6}{-10} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0.6 \:$

We know that magnification for a convex mirror is less then 1 .

Answer 2

Answer:

Explanation:

Given :-

u = - 10 cm

f = 15 cm

To Find :-

v = ?

m = ?

Formula to be used :-

Using mirror formula, 1/f = 1/v + 1/u

Solution :-

Putting all the values, we get

1/f = 1/v + 1/u

⇒ 1/ 15 = 1/v + 1/- 10

⇒ 1/v = 1/10 + 1/15

⇒ 1/v = 3 + 2/30

⇒ 1/v = 5/30

⇒ v = 6 cm

Now, Magnification, m = h’/h = - v/u

m = - v/u

⇒ m = - 6/ - 10

⇒ m = 0.6

Nature of the Image :-

Here, v is positive so the image is inverted and real.