Question

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm,The distance of the object fo

Answer 1

Correct Question -

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm .

The image Height is 4 cm .

Find the distance of the object from the pole of the mirror .

Solution -

Here , in the above question , the following information is given -

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm .

The image Height is 4 cm

To find -

Object distance .

According to the lens Formula -

$\sf{ \dfrac{ 1}{ f } = \dfrac{ 1}{ v } - \dfrac{1}{ u } }$

However , in a convex lens, u is always negative .

So ,

The new formula becomes -

$\sf{ \dfrac{ 1}{ f } = \dfrac{ 1}{ v } + \dfrac{1}{ u } }$

Substuting the given Values -

$\sf{ \dfrac{ 1}{ 20 } = \dfrac{ 1}{ v } + \dfrac{1}{ u } }$

Now we know that -

$\sf{ Magnification = \dfrac{ h }{ h_{0} } = \dfrac{ v }{ u } }$

Now , according to the given Values -

$\sf{ \dfrac{ v }{ u } = 2 } \\ \\ \sf{ \implies { v = 2u }}$

Substituting this into the first Equation -

$\sf{ \dfrac{ 1}{ 20 } = \dfrac{ 1}{ u } + \dfrac{1}{ 2u } }$

$\sf{ \implies { ( \dfrac{ 1}{ u } )( 1 + \dfrac{1}{2} ) = \dfrac{1}{20} } }$

$\sf{ \implies { \dfrac{3}{ 2u } = \dfrac{1}{ 20 } }} \\ \\ \sf{ \implies { u = 30 \: cm }}$

Hence the required object distance is 30 cm ..... A

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Answer 2

COMPLETE QUESTION

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find nature, position and size of image. Also, find it's magnification.

$\boxed{\rule{200}2}$

GiVeN :-

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm.

Object size, h$\sf{_o}$ = 2 cm

Focal length, f = 20 cm

Object distance, u = -15 cm

To DeTeRmInE :-

The image distance (v), magnification(M) and image size (h$\sf{_i}$).

SoLuTiOn :-

We know the lens formula :-

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

Putting the values :--

$\displaystyle{\sf{\dashrightarrow \frac{1}{v}-\frac{1}{-15}=\frac{1}{20} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}+\frac{1}{15}=\frac{1}{20} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{15} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{3-4}{60} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{-1}{60}}}\\\\\boxed{\boxed{\displaystyle{\color{orange}{\sf{\dashrightarrow v=-60\ cm}}}}}$

$\rule{100}2$

We also know,

$\boxed{\sf{\dag\ Magnification=\dfrac{v}{u} }}$

Putting the values :--

$\displaystyle{\sf{\dashrightarrow M=\frac{-60 }{-15} }}\\\\\boxed{\boxed{\color{orange}{\displaystyle{\sf{\dashrightarrow M=4\ cm}}}}}$

$\rule{100}2$

We also know,

$\boxed{\sf{\dag\ M=\dfrac{h_i}{h_o} }}$

Putting the values :--

$\displaystyle{\sf{\dashrightarrow 4=\frac{h_i}{2} }}\\\\\displaystyle{\sf{\dashrightarrow h_i=4\times2}}\\\\\boxed{\boxed{\displaystyle{\color{orange}{\sf{\dashrightarrow h_i=8\ cm}}}}}$

• Hence, the image is formed 60 in front of the mirror (i.e. to the left side).
• The magnification is 4, so the image is virtual and erect.
• The image is enlarged as hi > ho.