Question

an object of hieght 4cm is placed at 40cm from concave lens of focal length 20cm . calculate the position of the image and size of the image.


plz solve this. need urgently​

Answer 1

★ Figure refer to attachment

Given

An object of height 4cm is placed at 40cm from concave lens of focal length 20cm .

To find

calculate the position of the image and size of the image.

Solution

• height of object (ho) = +4cm
• Distance of object (u) = - 40cm
• focal length (f) = -20cm
• Image distance (v) = ?
• Height of image (hi) = ?

According to lens formula

$\implies\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{v}-\dfrac{-1}{40}=\dfrac{-1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1}{40}-\dfrac{1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1-2}{40} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-3}{40} \\ \\ \\ \implies\sf v=-13.3cm$

Now,

• v = -13.3cm
• ho = +4cm
• hi = ?

$\implies\sf m=\dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \\ \implies\sf \dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \\ \implies\sf \dfrac{-13.3}{-40}=\dfrac{h_i}{4} \\ \\ \\ \implies\sf h_i=\dfrac{4\times{13.3}}{40} \\ \\ \\ \implies\sf h_i=+1.3cm$

Nature of image

• Nature = Virtual and erect
• Size = Diminished
• Position = b/w F1 and optical centre

Hence, the position of image is -13.3cm and the size of image is +1.3cm

Answer 2

$\large \dag\ \underline{\mathbb{GIVEN :-}}$

$\textsf{An object of height 4 cm is placed at 40 cm from }\\\textsf{concave lens of focal length 20 cm.}$

$\bullet\ \sf{Size/Height\ of\ the\ object(h_o)=4\ cm}$

$\bullet\ \sf{Object\ distance(u)=-40\ cm}$

$\bullet\ \sf{Focal\ length(f)=-20\ cm}$

$\large \dag\ \underline{\mathbb{TO\ CALCULATE :-}}$

$\textsf{The position and the size of the image.}$

$\large \dag\ \underline{\mathbb{FORMULAS\ TO\ BE\ USED :-}}$

$\underline{\sf{\bigstar\ LENS\ FORMULA:-}}$

$\displaystyle{\sf{\frac{1}{v}-\frac{1}{u}=\frac{1}{f} }}$

$\underline{\sf{\bigstar\ MAGNIFICATION:-}}$

$\sf{m=\dfrac{v}{u}=\dfrac{h_i}{h_o} }$

$\large \dag\ \underline{\mathbb{SOLUTION:-}}$

$\textsf{Using the lens formula :-}$

$\displaystyle{\sf{\frac{1}{v}-\frac{1}{u} =\frac{1}{f} }}$

$\displaystyle{\sf{\longmapsto \frac{1}{v}=\frac{1}{-20}+\frac{1}{-40} }}\\\\\\\displaystyle{\sf{\longmapsto\ \frac{1}{v}=\frac{-1}{20}-\frac{1}{40} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{1}{v}= \frac{-2-1}{40} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{1}{v}=\frac{-3}{40} }}\\\\\\\boxed{\displaystyle{\sf{\longmapsto v=\frac{-40}{3} \ cm}}\\\\\\}$

$\rule{150}2$

$\sf{Now,}$

$\displaystyle{\sf{m=\frac{v}{u}=\frac{h_i}{h_o} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{\frac{-40}{3} }{-40}=\frac{h_i}{4} }}\\\\\\\displaystyle{\sf{m=\frac{v}{u}=\frac{h_i}{h_o} }}\\\\\\\displaystyle{\sf{\longmapsto \frac{1}{3}=\frac{h_i}{4} }}\\\\\\\displaystyle{\sf{m=\frac{v}{u}=\frac{h_i}{h_o} }}\\\\\\\displaystyle{\sf{\longmapsto h_i=\frac{4}{3} }}$

$\boxed{\displaystyle{\sf{\longmapsto h_i=1.33\ cm}}}$

$\underline{\textbf{Therefore, the image is -40/3 cm from the }}\\\underline{\textbf{lens and its size is 1.33 cm.}}$