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Question:- if an object of 7 cm height is placed at a distance of 12 m from the convex lens of the focal length 8 centimetre find the position nature and height of the image.

Answer 1

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Given –

object distance, u = - 12cm
image distance, v =?
focal length, f = 8 cm

using lens formula

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{8} = \frac{1}{v} - \frac{1}{ - 12} \\ \frac{1}{8} = \frac{1}{v} + \frac{1}{12} \\ \frac{1}{v} = \frac{1}{8} - \frac{1}{12} \\ \frac{1}{v} = \frac{3 - 2}{24} \\ \frac{1}{v} = \frac{1}{24} \\ v = 24$
using the formula of magnification
$m = \frac{v}{u}$
$m = \frac{24}{ - 12} \\ m = - 2$
the height of the image
$m = \frac{h2}{h1} \\ - 2 = \frac{h2}{7} \\ h2 = - 2 \times 7 \\ h2 = - 14$

Answer 2

 

Given – 

object distance, u = - 12cm
image distance, v =?
focal length, f = 8 cm

using lens formula

\begin{lgathered}\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\end{lgathered}f1​=v1​−u1​​ 
\begin{lgathered}\frac{1}{8} = \frac{1}{v} - \frac{1}{ - 12} \\ \frac{1}{8} = \frac{1}{v} + \frac{1}{12} \\ \frac{1}{v} = \frac{1}{8} - \frac{1}{12} \\ \frac{1}{v} = \frac{3 - 2}{24} \\ \frac{1}{v} = \frac{1}{24} \\ v = 24\end{lgathered}81​=v1​−−121​81​=v1​+121​v1​=81​−121​v1​=243−2​v1​=241​v=24​ 
using the formula of magnification
m = \frac{v}{u}m=uv​ 
\begin{lgathered}m = \frac{24}{ - 12} \\ m = - 2\end{lgathered}m=−1224​m=−2​ 
the height of the image
\begin{lgathered}m = \frac{h2}{h1} \\ - 2 = \frac{h2}{7} \\ h2 = - 2 \times 7 \\ h2 = - 14\end{lgathered}m=h1h2​−2=7h2​h2=−2×7h2=−14​