A coin 1.9 cm in diameter is held 40 cm from a concave spherical mirror of 30 cm radius of...

Question:

A coin 1.9 cm in diameter is held 40 cm from a concave spherical mirror of 30 cm radius of curvature. Locate the image of the coin Then what is the diameter of coin?

Image Formation:

A concave mirror is a type where the reflection portion of the mirror is the inside part of the curved structure. The image formation with the help of a concave mirror can be either a real image or a virtual image depending upon the object distance from the mirror.

Variables used are,

{eq}D \rightarrow \textrm{diameter of the coin}\\ D' \rightarrow \textrm{diameter of the image coin}\\ f \rightarrow \textrm{focal length of the mirror}\\ u \rightarrow \textrm{object distance from the mirror}\\ v \rightarrow \textrm{image distance}\\ m \rightarrow \textrm{magnification of the concave mirror} {/eq}

Note The radius of curvature is twice the measure of the focal length of the mirror.

Requisite Formula

• Mirror equation {eq}\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} {/eq}

Substituting the given values we get,

{eq}\dfrac{-1}{15}\;\rm cm = \dfrac{1}{v} + \dfrac{-1}{40}\;\rm cm\\ \dfrac{1}{v} = \dfrac{-1}{15}\;\rm cm + \dfrac{1}{40}\;\rm cm\\ v = \dfrac{-1}{0.042}\;\rm cm\\ \boxed{v = - 23.81\;\rm cm} {/eq}

The negative image distance conveys that the image is formed on the same side as the object position.

The mathematical expression for the magnification of a concave mirror is {eq}m = \dfrac{-v}{u}\\ m = -\dfrac{-23.81\;\rm cm}{-40\;\rm cm}\\ \boxed{m \approx - 0.59} {/eq}

The negative magnification exhibits that the image formed is diminished in size and the nature of the image is inverted and real (forming below the principal axis).

The diameter of the coin of the image will be (neglecting the negative sign),

{eq}D' = (0.59)(1.9)\\ \boxed{D' \approx 1.12\;\rm cm} {/eq}