(a) The power of a lens 'X' is -2·5 D. Name the lens and determine its focal length in cm. For which eye defect of vision will an optician prescribe this type of lens as a corrective lens ?

(b) "The value of magnification 'm' for a lens is -2." Using new Cartesian Sign Convention and considering that an object is placed at a distance of 20 cm from the optical centre of this lens, state :

1. the nature of the image formed;
2. size of the image compared to the size of the object;
3. position of the image, and
4. sign of the height of the image.

(c) The numerical values of the focal lengths of two lenses A and B are 10 cm and 20 cm respectively. Which one of the two will show higher degree of convergence/divergence? Give reason to justify your answer.

(a) Given,

• Power of the lens ($\text P$) = -2.5 D

As power of the lens is negative it means the lens is concave (diverging).

Power of a lens is given by,

$\text P = \dfrac{1}{\text {f (in m)}}$

Where $\text f$ is the focal length of the lens in metre.

$\Rightarrow \text f = \dfrac{1}{\text P} \\[1em] = \dfrac{1}{-2.5} \\[1em] = -\dfrac{1}{2.5} \\[1em] = -\dfrac{10}{25} \\[1em] = -\dfrac{2}{5} \\[1em] \Rightarrow \text f = -0.4\ \text m \\[1em] \Rightarrow \text f = -40 \text { cm}$

Since the lens is diverging in nature then such a lens is prescribed to correct myopia (short-sightedness).

(b) Given,

• Magnification ($\text m$) = -2
• Object distance ($\text u$) = -20 cm

1. Since m is negative it means image is real and inverted.

2. Since m = 2 so it means size of the image is two times the size of the object.

3. Let, the position of the image be $\text v$.

As, magnification of an object is given by,

$\text m = \dfrac{\text v}{\text u} \\[1em] \Rightarrow \text v = \text m \times \text u \\[1em] = (-2) \times (-20) \\[1em] \Rightarrow \text v = 40\ \text { cm}$

4. As, $\text m$ < 0 it means image is inverted which implies sign of the height of the image negative.

(c) Given,

• Focal length of lens A = 10 cm
• Focal length of lens B = 20 cm

As, power of a lens is given by,

$\text P = \dfrac{1}{\text {f (in m)}}$,

Hence, lens A will show a higher degree of convergence/divergence because it has a smaller focal length and therefore higher power.