Question

an object is stand in front of a mirror which produces two times magnified image on the screen at a distance of 30 cm calculate focal length of mirror and name the mirror also draw a ray diagram for given situation​

Answer 1

Given :

• Magnification = +2
• Imagery distance (v) = 30 cm

To FinD :

• Focal length of mirror

SolutioN :

Use formula for magnification :

⇒m = -v/u

⇒2 = -30/u

⇒1/u = -2/30

⇒1/u = -30/2

⇒u = -15

$\therefore$ Object is placed at distance of 15 cm in front of mirror

_______________________________

Now, use Mirror formula :

⇒1/f = 1/v + 1/u

⇒1/f = 1/30 + (-1/15)

⇒1/f = 1/30 - 1/15

⇒1/f = (1 - 2)/30

⇒1/f = -1/30

⇒f = - 30

$\therefore$ Focal length of mirror is - 30 cm.

• Mirror is concave mirror

Answer 2

Answer:

Given:

• An object is stand in front of a mirror which produces two times magnified image on the screen at a distance of 30 cm.

Find:

• Calculate focal length of mirror and name the mirror.

Know terms:

1. Magnification = (m)
2. Distance of image = (v)
3. Distance of object = (u)

Using formula:

★ ${\sf{\underline{\boxed{\orange{\sf{m = \dfrac{-v}{u} }}}}}}$

Calculations:

⇒ $\bold{2 = \dfrac{-30}{u}}$

⇒ $\bold{\dfrac{1}{u} = \dfrac{-2}{30}}$

⇒ $\bold{\dfrac{1}{u} = \dfrac{-30}{2}}$

⇒ ${\sf{\underline{\boxed{\red{\sf{u = -15}}}}}}$

Therefore, 15 cm is the distance of the object at the front of the mirror.

Using formula:

★ ${\sf{\underline{\boxed{\orange{\sf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }}}}}}$

Calculations:

⇒ $\bold{\dfrac{1}{f} = \dfrac{1}{30} + (\dfrac{-1}{15})}$

⇒ $\bold{\dfrac{1}{f} = \dfrac{1}{30} - \dfrac{1}{15}}$

⇒ $\bold{\dfrac{1}{f} = \dfrac{1 - 2}{30}}$

⇒ $\bold{\dfrac{1}{f} = \dfrac{-1}{30}}$

⇒ ${\sf{\underline{\boxed{\red{\sf{f = -30 \: cm }}}}}}$

Therefore, -30 cm is the focal length of mirror.

The mirror is concave mirror.

Note:

• Refer the attachment for Ray diagram.