Question

An object 4.5 cm high is placed at a distance of 12 cm from a concave lens of focal length 18 cm.find the size of the image.​

Answer 1

Answer:

the answer is 50 size if the image...

Hope its help..

Answer 2

$\tt \blue{ Solution:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{ \:Height \: of \: object(ho) = 4.5cm}$

$\footnotesize \tt{ Distance \: of \: object(u) = 12cm}$

$\footnotesize \tt{ Focal \: length(f)= 18cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\hline \underline{\begin{array}{ ||c||c|| } \underline {\footnotesize \tt{ In \: Concave \: lens }}& \underline{\footnotesize \tt{Value} }\\\footnotesize \tt{u }& \footnotesize \tt{negative \: value}\\\footnotesize \tt{v}& \footnotesize \tt{negative \: value}\\\footnotesize \tt{f}& \footnotesize \tt{negative \: value}\end{array}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{ Formula \: Used:- }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }$

$\tt{ \frac{1}{ - 18} = \frac{1}{v} - \frac{1}{ - 12} }$

$\tt{ \frac{1}{ - 18} = \frac{1}{v} + \frac{1}{ 12} }$

$\tt{ - \frac{1}{ 18} - \frac{1}{12} = \frac{1}{v} }$

$\tt{ \frac{ - 2 \: - \: 3}{ 36} = \frac{1}{v} }$

$\tt{ \frac{ - 5}{ 36} = \frac{1}{v} }$

$\footnotesize\tt{ - 5v = 36 }$

$\footnotesize\boxed{ \underline{\tt{ Distance \: of \: Image(v) = \frac{ - 36}{5} }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{Magnification (m) \: } \tt{= \: \frac{ v}{u} }$

$\footnotesize \tt{Magnification (m) \: } \tt{= \: \frac{ - 36}{5 \times ( - 12)} }$

$\footnotesize \tt{Magnification (m) \: } \tt{= \: \frac{ \cancel{- 36}}{5 \times ( - \cancel {12})} }$

$\footnotesize\boxed{ \underline{\footnotesize \tt{Magnification (m) \: } \tt{= \: \frac{ 3}{5} }}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize \tt{Magnification (m) \: } \tt{= \: \frac{ hi}{ho} }$

$\footnotesize \tt{ \frac{3}{5} \: } \tt{= \: \frac{ hi}{4.5} }$

$\footnotesize \tt{ 3 \times 4.5}\tt{ \: = 5hi }$

$\footnotesize \tt{ 13.5}\tt{ \: = 5hi }$

$\boxed{ \underline{ \footnotesize \tt{ 2.7 }\tt{ \: = hi (Height \size \: of \: Object)}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\small\tt{ Important \: tables:- }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\hline \underline{\begin{array}{ ||c||c|| } \underline {\footnotesize \tt{ Image \: formed }}& \underline{\footnotesize \tt{Value} }\\\footnotesize \tt{Real \: and \: Inverted}& \footnotesize \tt{m = negative \: value}\\\footnotesize \tt{Virtual \: and \:Erect}& \footnotesize \tt{m = positive \: value} \\\footnotesize \tt{Enlarged \: size}&\footnotesize \tt{m > 1 }\\\footnotesize \tt{Same \: size}&\footnotesize \tt{m = 1} \\\footnotesize \tt{Diminished \: size}&\footnotesize \tt{m < 1 \:}\end{array}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\hline \underline{\begin{array}{ ||c||c|| } \underline {\footnotesize \tt{ In \: Concave \: lens }}& \underline{\footnotesize \tt{Image \: formed} }\\\footnotesize \tt{v \: = \: positive \: value}& \footnotesize \tt{behind \: the \: lens}\\\footnotesize \tt{v =negative \: value }& \footnotesize \tt{ \: front \: of \: lens}\end{array}}$