Question

A 5 cm tall object is placed in front of a convex lens of focal length 20 cm at a distance 30 cm from the lens. Find the position, size of the image and magnification. 3 points

Answer 1

★ Figure refer to attachment

$\Large{\bf{\underline{\underline{\blue{Given}}}}}$

• Height of the object (ho) = + 5cm
• Focal length (f) = + 20cm
• Object distance (u) = - 30cm

$\Large{\bf{\underline{\underline{\blue{Find\:out}}}}}$

• Find the position, size of the image and magnification

$\Large{\bf{\underline{\underline{\blue{Solution}}}}}$

Apply lens formula

$\implies\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1}{30}+\dfrac{1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-2+3}{60} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{1}{60} \\ \\ \\ \implies\sf v=+60cm$

• Image distance = +60cm

Now,

$\implies\sf m=\dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \\ \implies\sf \dfrac{v}{u}=\dfrac{h_i}{h_o} \\ \\ \\ \implies\sf \dfrac{-60}{30}=\dfrac{-h_i}{5} \\ \\ \\ \implies\sf -2=\dfrac{h_i}{5} \\ \\ \\ \implies\sf h_i=-10cm$

• Size of image = -10cm

________________________________

$\implies\sf Magnification =\dfrac{v}{u} \\ \\ \\ \implies\sf m=\dfrac{60}{-30} \\ \\ \\ \implies\sf m=-2$

• Magnification = -2

_________________________

• Nature of image
• Nature = Real and inverted
• Size = Enlarged
• Position = Beyond 2F₂

Answer 2

→ANSWER←

• Position : Beyond C₂
• Magnification : -2
• Size : Enlarged
• Nature : Real & inverted

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GIVEN

A 5 cm tall object is placed in front of a convex lens of focal length 20 cm at a distance 30 cm from the lens.

$\bullet\ \sf{h_o=5\ cm}$

$\bullet\ \sf{f=20\ cm}$

$\bullet\ \sf{u=-30\ cm}$

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TO FIND

The position, size and the magnification of the image.

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SOLUTION

We know the LENS FORMULA :-

$\boxed{\sf{\ddag\:\: \frac{1}{v}-\frac{1}{u}=\frac{1}{u} }}$

Substituting the known values :--

$\displaystyle{\sf{\dashrightarrow \frac{1}{v}-\frac{1}{-30} = \frac{1}{20} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}+\frac{1}{30}=\frac{1}{20} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{30} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{3-2}{60} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{1}{60} }}\\\\\boxed{\displaystyle{\sf{\dashrightarrow v=60\ cm}}}$

→ Therefore, the image is formed 60 cm on the right side of the lens.(Beyond the C₂)

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We also know,

$\boxed{\displaystyle{\sf{\ddag\:\: Magnification=\frac{v}{u} }}}$

Substituting the known values :-

$\displaystyle{\sf{\dashrightarrow m=\frac{60}{-30} }}\\\\\boxed{\displaystyle{\sf{\dashrightarrow m=\frac{-2}{1} =-2}}}$

→ The magnification is -2.

→ Therefore, the image is real and inverted for the magnification is negative.

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As we know that,

$\boxed{\displaystyle{\sf{\ddag\:\: \frac{v}{u}=\frac{h_i}{h_o} }}}$

Substituting the known values :-

$\displaystyle{\sf{\dashrightarrow \frac{60}{-30}=\frac{h_i}{5} }}\\\\\displaystyle{\sf{\dashrightarrow -2=\frac{h_i}{5} }}\\\\\displaystyle{\sf{\dashrightarrow h=-2\times5}}\\\\\boxed{\displaystyle{\sf{\dashrightarrow h_i=-10\ cm}}}$

→ h$\sf{_i}$ > h$\sf{_o}$ in magnitude. So the image is enlarged.