Zarina worked as an apprentice in a factory where flashlights and solar cookers are made. She learnt to make the circuits, the design of the light-box and light concentrators of the solar cookers as well. She learnt the uses of lenses in making all those tools. Based on your understanding of lenses, answer the following questions.

(a) What kind of lenses are used in the flashlight and light concentrator of the solar-cooker?

(b) Give reasons for your choices in your answer for part A.

Attempt either subpart C or D.

(c) An object is placed 40 cm away from a lens which is normally used in a solar-cooker. The image formed is twice the size of the object. Calculate the focal length of the lens.

OR

(d) An object is placed 20 cm in front of a lens which is used in a flashlight, and the image is formed 10 cm away from the lens on the same side as the object. Calculate the focal length of the lens.

(a) Concave Lens for Flashlight and Convex Lens for solar cooker.

(b) Concave lens diverges the light rays which is needed for a wider reach of the flashlight and convex lens converges the rays which helps to raise the temperature of the place where rays converge.

(c) Given,

• Object distance ($\text u$) = -40 cm
• Size of the image ($\text I$) = -2 x Size of the object ($\text O$)

Here, negative sign indicates that the object is in front of the lens and the image is inverted.

As, magnification ($\text m$) of a lens is given by,

$\text m = \dfrac{\text I}{\text O} \\[1em] = \dfrac{-2\times \text O}{\text O} \\[1em] = -2 \\[1em]$

Let, image distance be '$\text v$'.

Then,

$\text m = \dfrac{\text v}{\text u} \\[1em] \Rightarrow \text v = \text m\text u \\[1em] = -2 \times (-40) \\[1em] \Rightarrow \text v = 80\ \text {cm}$

Let, focal length of the lens be '$\text f$'.

Then, by using the lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1}{80} - \dfrac{1}{(-40)} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1}{80} + \dfrac{1}{40} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1 + 2}{80} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{3}{80} \\[1em] \Rightarrow \text f = \dfrac{80}{3} \\[1em] \Rightarrow \text f \approx 26.67 \text { cm}$

Hence, the focal length of the lens is approximately 26.67 cm.

(d) Given,

• Object distance ($\text u$) = -20 cm
• Image distance ($\text v$) = -10 cm

Here, negative sign indicates that both the object and the image are in front of the lens.

Let, focal length of the lens be '$\text f$'.

Then, by using the lens formula,

$\dfrac{1}{\text f} = \dfrac{1}{\text v} - \dfrac{1}{\text u} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1}{(-10)} - \dfrac{1}{(-20)} \\[1em] \Rightarrow \dfrac{1}{\text f} = -\dfrac{1}{10} + \dfrac{1}{20} \\[1em] \Rightarrow \dfrac{1}{\text f} = \dfrac{1 - 2}{20} \\[1em] \Rightarrow \dfrac{1}{\text f} = -\dfrac{1}{20} \\[1em] \Rightarrow \text f = -20 \text { cm}$

Hence, the focal length of the lens is -20 cm, indicating it is a diverging lens (concave lens).