Question

A square card of side length 1 mm is being through a magnifying lens of focal length 10 cm. The card is placed at a distance of 9 cm from the lens. The apparent area of the card through the lens is

A.)$1c{m^2}$
B.)$0.81c{m^2}$
C.)$0.27c{m^2}$
D.)$0.60c{m^2}$

Answer 1

Hint- In order to solve these types of questions, we will use the concept of power of the lens and we will proceed further first by calculating the focal length using the lens formula and then proceed further for finding the power of the lens.

Complete step-by-step answer:
Formula used:

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Where f is the focal length, v is the image distance and u is the object distance.
Given

Area of the square card = $1mm \times 1mm = 1m{m^2}$
Focal length of magnifying lens, ${\text{f}} = + 10cm$
Object distance u = -9 cm

Let f be the focal length, v be the image distance and u be the object distance.

As we know $\left[ {\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}} \right]$
$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

Substituting the value of u and v from the given conditions in the above equation

$
  \dfrac{1}{v} = \dfrac{1}{{ + 10}} + \dfrac{1}{{ - 9}} \\
  \dfrac{1}{v} = \dfrac{1}{{ - 90}} \\
  v = - 90cm \\
$

Magnification, $m = \dfrac{v}{u} = \dfrac{{ - 90cm}}{{9cm}} = 10$

Therefore the apparent area of the card through the lens
$
   = 10 \times 10 \times 1m{m^2} = 100m{m^2} \\
   = 1c{m^2} \\
 $

Hence, the correct option is A.

Additional Information- Lenses can be classified into two major forms of large measure: convex and concave. The lenses in the middle are smoother than the corners, while the lenses along the sides are concave. The mirror concentrates on a spot on the other side of the frame, a light ray traveling through a convex prism. This is considered the focus of publicity. In the case of concave lenses that diverge from the condensation of light waves, the focal point is on the source of incoming light that appears to come from the direction of light through the lens.

Note- In order to solve these types of questions, apart from the formulas you must remember the sign conventions. Some of the sign conventions are- for convex lens the focal length is positive and for concave lens the focal length is negative. Similarly, remember the other conventions. Most optical devices make use of not just one lens, but of a combination of convex and concave lenses.