Question

Part I
1.
An object of size 3.0 cm is placed at a distance of 14 cin from a concave lens of focal length 21 cm,
Find the nature, position and size of the image. What happens if the object is moved further from
the lens?
c​

Answer 1

GiveN :-

• An object of size 3.0 cm is placed at a distance of 14 cm from a concave lens of focal length 21 cm .

To FinD :-

• The nature, position and size of the image.
• What happens if the object is moved further fromthe lens?

SolutioN :-

Given that a object of 3cm is placed in front of s concave lens of focal length 21cm .We need to find the position and size of the image . Here we can use the lens formula .

★ Using Len's Formula :-

$\sf:\implies \pink{\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}}\\\\\sf:\implies \dfrac{1}{v}-\dfrac{1}{-14cm}=\dfrac{1}{-21 cm} \\\\\sf:\implies \dfrac{1}{v}=\dfrac{1}{14cm}-\dfrac{1}{21cm}\\\\\sf:\implies \dfrac{1}{v}=\dfrac{3-2}{42cm}\\\\\sf:\implies\dfrac{1}{v}=\dfrac{1}{42cm}\\\\\sf:\implies\boxed{\pink{\frak{ Image\ Distance \ =\ 42\ cm .}}}$

$\rule{200}2$

★ Using formula of Magnification :-

$\sf:\implies \pink{ M =-\dfrac{ Image \ Distance}{Object\ Distance } }\\\\\sf:\implies \dfrac{H_{(image)}}{H_{(Object)}}=\dfrac{-42cm}{-14cm}\\\\\sf:\implies \dfrac{H_i}{3cm}= 3 \\\\\sf:\implies \boxed{\pink{\frak{ Image\ Height \ =\ 9\ cm .}}}$