Question

(a) Find the nature, position and magnification of the images formed by a convex lens of focal length 0.20 m if the object is placed at a distance of:
(i) 0.50 m
(ii) 0.25 m
(iii) 0.15 m

(b) Which of the above cases represents the use of convex lens in a film projector, in a camera, and as a magnifying glass?

Answer 1

(i) u = $-$0.50 m = $-$50 cm
f = 0.20 m = 20 cm
Lens formula:
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{20}=\frac{1}{v}-\frac{1}{-50} \\ \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{50} \\ \Rightarrow \frac{1}{v}=\frac{5-2}{100} \\ \Rightarrow \frac{1}{v}=\frac{3}{100} \\ \Rightarrow v=\frac{100}{3}\mathrm{cm} \\ \mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}33.3\mathrm{cm}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{mirror} \\ \mathrm{magnification}m=\frac{v}{u} \\ m=\frac{33.3}{-50} \\ m=-0.66 \\ \mathrm{magnification}\mathrm{is}\mathrm{negative}\mathrm{therefore}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\mathrm{inverted}. \end{gathered}$$

(ii) u = $-$25 cm
f = 20 cm
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{20}=\frac{1}{v}-\frac{1}{-25} \\ \Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{25} \\ \Rightarrow \frac{1}{v}=\frac{5-4}{100} \\ \Rightarrow \frac{1}{v}=\frac{1}{100} \\ \Rightarrow \mathrm{v}=100\mathrm{cm} \\ \mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}100\mathrm{cm}\mathrm{behind}\mathrm{the}\mathrm{mirrror}. \\ \mathrm{Magnification},\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\ \Rightarrow \mathrm{m}=\frac{100}{-25} \\ \Rightarrow \mathrm{m}=-4 \\ \mathrm{Magnification}\mathrm{is}\mathrm{negative},\mathrm{which}\mathrm{means}\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{real}\mathrm{and}\mathrm{inverted}. \end{gathered}$$

(c) u = -15 cm
f = 20

$$\begin{gathered} \mathrm{Substituting}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{lens}\mathrm{formula},\mathrm{we}\mathrm{get}: \\ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{20}=\frac{1}{v}-\frac{1}{-15} \\ \Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{15} \\ \Rightarrow \frac{1}{20}-\frac{1}{15}=\frac{1}{v} \\ \Rightarrow \frac{1}{v}=\frac{3-4}{60} \\ \Rightarrow \frac{1}{v}=-\frac{1}{60} \\ v=-60\mathrm{cm} \\ \mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}60\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{mirror}. \\ m=\frac{v}{u} \\ \Rightarrow m=\frac{-60}{-15} \\ \Rightarrow m=4 \\ \mathrm{Magnification}\mathrm{is}\mathrm{positive};\mathrm{therefore},\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{erect}. \end{gathered}$$

(b)
We can use case (ii) for a film projector.
We can use case (i) for a camera.
We can use case (ii) for a magnifying glass.