Question

Under which of the following conditions of the objective distance a concave mirror of focal length 40 cm can form a virtual image larger than the actual object?
A. – 30 cm
B. – 80 cm
C. – 40 cm
D. – 60 cm

Answer 1

The mirror formula along with the magnification formula can be used to solve this problem. Plotting a ray diagram is the best solution to understand the given problem clearly. When the object is placed between the focus and the pole, an enlarged, erect and virtual image gets formed behind the mirror.
Formulae used:
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
$m=\left( \dfrac{v}{u} \right)$

Complete answer:
From given, we have the data,
The focal length of the concave mirror = 40 cm.
A virtual image is a collection of focal points made by extensions of diverging rays. In simple words, a virtual image is opposite to that of a real image. An image that cannot be obtained on the screen is the virtual one.
The ray diagram of virtual image formation is as follows.

Where C represents the centre of curvature, F represents the focus and P represents the pole.
Bold arrow represents the object and the dotted arrow represents the virtual image.
When the position of the object is between the focus and the pole, an enlarged, virtual image forms.
The mirror formula for the concave mirror is given as follows:
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
Where f is the focal length, u is the object distance and v is the image distance.
$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$ …… (1)
For a concave mirror, the focal length, f is negative f < 0
Also, the object distance, u is negative, u < 0
This implies that,
$\dfrac{1}{f}-\dfrac{1}{u}>0$
Therefore, equation (1) becomes
$\dfrac{1}{v}>0$
The above equation implies that the image formed is on the right side and is virtual.
The magnification formula in the case of a concave mirror for producing a virtual image is,
$m=-\left( \dfrac{v}{u} \right)$
Where v is the image distance and u is the object distance
The value of magnification should be greater than one, in the case of enlarged image formation.
$m>1$
$m=\dfrac{\left| f \right|}{\left| f \right|-\left| u \right|}>1$
Substitute the value of the focal length given.

& \dfrac{40}{40-\left| u \right|}>1 \\\ & \Rightarrow 40>40-\left| u \right| \\\ & \Rightarrow \left| u \right|>0 \\\ \end{aligned}$$ The object placed between the focus and the pole of a concave mirror produces a virtual and enlarged image. As the value of the objective distance $$\left| -30 \right|$$ cm lies within the distance between the focus and the pole. **Thus, option (A) - 30 cm is correct.** **Additional information:** Concave mirrors can produce both real and virtual images. The images formed can be enlarged, reduced, or the same size as the object. The images can be inverted in the case of real images. The images can be upright in the case of virtual images. **Note:** The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration while solving the problem.