Question

(a) An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image.
(b) If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
(c) Which of the above two cases illustrates the working of a magnifying glass?

Answer 1

(a) Given:
Object distance (u) = $-$24
Focal length (f) = +8
Object height (h) = 3
$$\begin{gathered} \mathrm{Lens}\mathrm{formula}\mathrm{is}\mathrm{given}\mathrm{by}: \\ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{8}=\frac{1}{v}-\frac{1}{-24} \\ \Rightarrow \frac{1}{8}=\frac{1}{v}+\frac{1}{24} \\ \Rightarrow \frac{1}{8}-\frac{1}{24}=\frac{1}{v} \\ \Rightarrow \frac{3-1}{24}=\frac{1}{v} \\ \Rightarrow \frac{2}{24}=\frac{1}{v} \\ \Rightarrow v=12\mathrm{cm} \\ \mathrm{Image}\mathrm{will}\mathrm{be}\mathrm{form}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}12\mathrm{cm}\mathrm{on}\mathrm{the}\mathrm{right}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{convex}\mathrm{lens}. \\ \mathrm{Magnification}m=\frac{v}{u} \\ \Rightarrow m=\frac{12}{-24} \\ \Rightarrow m=-\frac{1}{2} \\ \mathrm{So},\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{diminished}. \\ \mathrm{Negative}\mathrm{value}\mathrm{of}\mathrm{magnification}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{real}\mathrm{and}\mathrm{inverted}. \end{gathered}$$
$$\begin{gathered} \mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}} \\ -\frac{1}{2}=\frac{{\mathrm{h}}_{\mathrm{i}}}{3} \\ {\mathrm{h}}_{\mathrm{i}}=-\frac{3}{2} \\ {\mathrm{h}}_{\mathrm{i}}=-1.5\mathrm{cm} \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}1.5\mathrm{cm}.\mathrm{Here},\mathrm{negative}\mathrm{sign}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{in}\mathrm{the}\mathrm{downward}\mathrm{direction}. \end{gathered}$$

(b) Object distance (u) = $-$3
$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \Rightarrow \frac{1}{8}=\frac{1}{v}-\frac{1}{-3} \\ \Rightarrow \frac{1}{8}=\frac{1}{v}+\frac{1}{3} \\ \Rightarrow \frac{1}{8}-\frac{1}{3}=\frac{1}{v} \\ \Rightarrow \frac{3-8}{24}=\frac{1}{v} \\ \Rightarrow \frac{-5}{24}=\frac{1}{v} \\ \Rightarrow v=-\frac{24}{5} \\ \Rightarrow v=-4.8\mathrm{cm} \\ \mathrm{The}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{of}4.8\mathrm{cm}\mathrm{in}\mathrm{front}\mathrm{of}\mathrm{the}\mathrm{lens}. \\ \mathrm{Magnification}\left(m\right)=\frac{v}{u} \\ \Rightarrow m=\frac{-4.8}{-3} \\ \Rightarrow m=1.6 \\ \mathrm{Positive}\mathrm{value}\mathrm{of}\mathrm{magnification}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{is}\mathrm{virtual}\mathrm{and}\mathrm{erect}. \\ \mathrm{m}=\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}} \\ \Rightarrow 1.6=\frac{{\mathrm{h}}_{\mathrm{i}}}{3} \\ \Rightarrow {\mathrm{h}}_{\mathrm{i}}=3\times 1.6=4.8\mathrm{cm} \\ \mathrm{Positive}\mathrm{sign}\mathrm{of}\mathrm{the}\mathrm{image}\mathrm{shows}\mathrm{that}\mathrm{the}\mathrm{image}\mathrm{will}\mathrm{be}\mathrm{formed}\mathrm{above}\mathrm{the}\mathrm{principal}\mathrm{axis}. \end{gathered}$$

(c) Case (b) illustrates the working of magnifying lens as the object is between the focus and optical centre.