What is the magnification produced by a concave lens of focal length 10 cm, when an image is formed at a distance of 5 cm from the lens?

CONCEPT:

• For a spherical lens,

\(\frac{1}{f} = \frac{1}{v}-\frac{1}{u}\;\;\;\;\; \ldots \left( 1 \right)\)

Where, f = Focal length, v = Image distance optical center, u = Object distance from the optical center

• The magnification  produced by a concave lens,

m = v/u

• In the case of a concave lens, if the object is at any position except infinity, the image will always form between the second focus and the optical center.
• The image formed will always be virtual, erect, and diminished in this case.
• So while putting the value of v in the above formula, we have to put it with the negative sign (As per the sign convention, the image will be in the opposite direction of the incident ray with respect to the optical center)
• In a similar manner, f will also be taken as negative for the concave lens.

CALCULATION:

Given: f = -10 cm (Concave lens focus is negative), v = - 5 cm

• Using equation 1,

\(⇒\frac{1}{-10} = \frac{1}{-5}-\frac{1}{u} \)

⇒ u = - 10 cm

• The magnification  produced by a concave lens,

⇒ \(m=\frac{-5}{-10}=\frac 12\)

= 1/2 < 1 (Image is virtual, erect (Upright), small in size)

• The image is half size of the object.

The correct answer is option 3