Question

Q.5 An object is placed at a distance of 12cm in front of a concave mirror .it forms a real image four times larger than the object , calculate the distance of the image from the mirror. Required question

Answer 1

Question :–

An object is placed at a distance of 12cm in front of a concave mirror it forms a real image four times larger than the object, calculate the distance of the image from the mirror.

Given :–

• u = 12cm
• It forms a real image four times larger than the object.

To Find :–

• Calculate the distance of the image from the mirror.

Solution :–

Here we use magnification formula :–

→ $\frac{ h_{o}}{ h_{i} } = \frac{ - v}{u}$

u = -12 [ it is -ve because the object is real ]

→ $\frac{ h_{o}}{ h_{i} } = \frac{4}{1}$

→ $\frac{ - v}{u} = \frac{4}{1}$

→ $\frac{ - v}{ - 12} = \frac{4}{1}$

→ $\frac{ \cancel- v}{ \cancel- 12} = 4$

→ $v = 4 \times 12$

→ $v = 48$

Hence,

The distance of the image from the mirror is 48 cm

And the nature of the image is :–

1. Virtual.
2. Large in size.
3. Erect.

Answer 2

Answer:

Given :–

u = 12cm

It forms a real image four times larger than the object.

To Find :–

Calculate the distance of the image from the mirror.

Solution :–

Here we use magnification formula :–

→ \frac{ h_{o}}{ h_{i} } = \frac{ - v}{u}

h

i

h

o

=

u

−v

u = -12 [ it is -ve because the object is real ]

→ \frac{ h_{o}}{ h_{i} } = \frac{4}{1}

h

i

h

o

=

1

4

→ \frac{ - v}{u} = \frac{4}{1}

u

−v

=

1

4

→ \frac{ - v}{ - 12} = \frac{4}{1}

−12

−v

=

1

4

→ \frac{ \cancel- v}{ \cancel- 12} = 4

−

12

−

v

=4

→ v = 4 \times 12v=4×12

→ v = 48v=48

Hence,

The distance of the image from the mirror is 48 cm

And the nature of the image is :–

Virtual.

Large in size.

Erect