Question

A concave mirror is designed so that a person 20.0 cm in front of the mirror sees and upright image magnified by a factor of two. What is the radius of curvature of this mirror?

Answer 1

Given:

The mirror is concave.

The distance of the image is

`d_0=20\text{ cm}`
`d_0=20\text{ cm}`

The image is magnified by a factor of two.

`m=2`

Required: the radius of curvature of the concave mirror.

Step-by-step explanation:

we know that the magnification is given by

`m=-(d_i)/(d_0)`

substitute the value of d0 and m in the above equation, and we get:

`\begin{gathered} 2=-\frac{d_i}{-20.\text{ cm}} \\ d_i=40\text{ cm} \end{gathered}`

now, we will calculate the focal length of the mirror.

the mirror formula is,

`(1)/(f)=(1)/(d_i)+(1)/(d_0)`

Plugging all the values in the above, we get:

`\begin{gathered} (1)/(f)=\frac{1}{40\text{ cm}}+\frac{1}{20\text{ cm}} \\ (1)/(f)=(1+2)/(40) \\ f=(40)/(3) \\ f=\text{ 13.33 cm} \end{gathered}`

now, we calculate the radius of curvature.

we know that,

`f=(R)/(2)`

Substitute the value of f in the above relation, we get:

`\begin{gathered} 13.33\text{ cm=}(R)/(2) \\ R=2*13.33\text{ cm} \\ R=26.66\text{ cm} \end{gathered}`

Thus, the radius of the curvature is

`26.66\text{ cm}`