A concave mirror produces a real image that is 3 times the size of an object placed in fro… — JAMB UTME

A concave mirror produces a real image that is $3$ times the size of an object placed in front of it. If the distance between the object and its image is $40\text{ cm}$ , what is the focal length of the mirror in $\text{cm}$ ?

Answer: 15 cm

Answer

The focal length of the concave mirror is

15\text{ cm}

Using the magnification relation

v = 3u

and the object-image separation of

40\text{ cm}

, we obtain

3u - u = 40\text{ cm}

, which yields

u = 20\text{ cm}

and

v = 60\text{ cm}

. Substituting these distances into the mirror equation

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

gives

\frac{1}{f} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

, so

f = 15\text{ cm}

Step-by-Step Solution

Relate the image distance

v

to the object distance

u

using the linear magnification formula

v = 3u

Since the mirror forms a real, magnified image that is 3 times the size of the object, linear magnification

m = \frac{v}{u} = 3

Formulate an equation from the given object-to-image separation distance to solve for

u

and

v

u = 20\text{ cm}

and

v = 60\text{ cm}

The separation distance between the image and object is

v - u = 40\text{ cm}

. Substituting

v = 3u

yields

2u = 40\text{ cm} \implies u = 20\text{ cm}

and

v = 60\text{ cm}

Substitute the values of

u

and

v

into the mirror formula to compute the focal length

f

f = 15\text{ cm}

Applying

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

gives

f = 15\text{ cm}

Key Concept

Linear magnification and mirror formula for concave mirrors

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