Question

Example 3:
Find the position and nature of the image of an object of height 3 cm, when placed 60 cm from a convex
mirror of focal length 15 cm​

Answer 1

Answer:

position : between f and c

nature : 1. : real image

2. : inverted

3. : diminished

Answer 2

$\star \; {\underline{\boxed{\orange{\pmb{\textbf{\textsf{ Given \;}}}}}}}$

$\begin{gathered} \\\end{gathered}$

• u = -60cm

• f = 15

$\begin{gathered} \\\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\star \;{\underline{\boxed{\purple{\pmb{\textbf{\textsf{ To\;Find \; :- }}}}}}}\end{gathered}$

$\begin{gathered} \\\end{gathered}$

• v = ?

$\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}$

$\begin{gathered} \\\end{gathered}\star \;{\underline{\boxed{\red{\pmb{\textbf{\textsf{ SoluTion \; :- }}}}}}}$

$\begin{gathered} \\ \\\end{gathered}$

❍ Formula Used

${\underline{\boxed{\pmb{\sf{ \: \sf \: Mirror \: formula = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{5}}}}}}$

$\begin{gathered}\begin{gathered} \\ \end{gathered}\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\dag \;{\underline{\underline{\sf{ \; Calculating \; the \; Distance}}}}\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ \frac{1}{v} = \frac{1}{60} + \frac{1}{15} }\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ \frac{1}{v} = \frac{5}{60 } }\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ \frac{1}{v} = \frac{5}{60 } }\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ \frac{1}{v} = \frac{1}{12 } }\end{gathered}$

$\begin{gathered} \\ \qquad \; \dashrightarrow \sf{ v \: = \: 12 }\end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \;{\underline{\boxed{\red{\pmb{\frak { Image Distance=12 }}}}}} \; \bigstar \\ \\ \end{gathered}$

$\begin{gathered}\begin{gathered} \\ \end{gathered}\end{gathered}\begin{gathered}\begin{gathered} \\ \\\end{gathered}\dag \;{\underline{\underline{\sf{ \; Calculating \; the \; Magnification}}}}\end{gathered}$

$\begin{gathered} \sf \implies{m = - \dfrac{12}{ - 60} } = \dfrac{ h_{2}}{ 3 } \\ \\ \implies \sf\dfrac{12}{60} = \frac{ h_{2} }{3} \\ \\ \implies \sf h_{2} = \dfrac{12}{60} \times 3 \\ \\ \sf \implies h_{2} = \frac{3}{5} \\ \\ \sf \implies h_{2} = 0.6\end{gathered} \\ \\ \begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\red{\pmb{\frak { Magnification = 0.6cm}}}}}} \; \\ \\ \end{gathered}$

$\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}$

∴ Virtual and erect image of size 0.6 cm will be formed at a distance of 12 cm behind the mirror .

$\begin{gathered}\begin{gathered} \\ \qquad{\rule{200pt}{2pt}} \end{gathered}\end{gathered}$