Question

An object is 24cm away from a concave mirror and its images l is 16cm form the mirror . Find the focul length and radius of curvature of the mirror , and the magnification of the image​

Answer 1

Answer:

The focal length is - 9.6  cm and Radius of curvature of the concave mirror is - 19.2 cm and the magnification of the image is - 2/3.

Explanation:

Object distance (u) = - 24 cm

Image distance (v) = - 16 cm

Using Mirror Formula:

$= \frac{1}{v} +\frac{1}{u} = \frac{1}{f}$

$=\frac{1}{-16}+\frac{1}{-24}$

$= -6 + \frac{- 4}{96}$

$= \frac{-10}{96}$

Focal length of the concave mirror = - 9.6  cm (Negative sign indicates that the image formed is real and inverted)

To find radius of curvature = (2) (Focal length of the concave mirror)

Radius Of Curvature of the Concave Mirror = - 19.2  cm

Magnification of concave mirror = $\frac{-v}{u}$

Magnification = $\frac{-16}{24}$

Simplifying further we get:

$=\frac{-8}{12}$

$=\frac{-4}{6}$

$Magnification =\frac{-2}{3}$

Therefore the focal length is - 9.6  cm and Radius of curvature of the concave mirror is - 19.2 cm and the magnification of the image is - 2/3.

Answer 2

Answer:

Object Distance ( u ) = 24 cm

Image Distance ( v ) = 16 cm

Focal Length = ?

Radius of Curvature = ?

Magnification = ?

According to Mirror Formula,

• 1/v + 1/u = 1/f

According to Concave Mirror Convention,

• u = -24 cm
• v = -16 cm

Substituting them in the formula we get,

⇒ -1/24 - 1/16 = 1/f

Taking LCM we get,

⇒ ( -2 - 3 )/ 48 = 1/f

⇒ -5/48 = 1/f

⇒ f = -48/5 = -9.6 cm

Here '-' is just a convention.

Radius of Curvature = 2 × f

⇒ Radius of Curvature = 2 × 9.6 = 19.2 cm

Magnification = -v/u

⇒ Magnification = - ( -16 )/( -24 )

⇒ Magnification = 16/(-24) = - 2/3 = -0.67

'-' indicates that the image is diminished.

Here the image formed is real and diminished