Question

An object 1 cm in height is placed at a distance of 0.80m from a double convex lens of focal length 0.2m. Find the location and size of the image formed by(a) calculation and (b) the graphical method.

Answer 1

Given :

➩ In Convex Lens

• Object Height = 1cm
• Object is placed at distance (u) = -0.8m or -80cm
• Focal lenght (f) = 0.2m or 20cm

Sign Convention

Object is placed at left side of lens therefore it is -ve

To Find :

• Location of image
• Size of image

Formula Used :

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

$\bullet\underline{\boxed{\sf Magnification (m)=\dfrac{h_i}{h_o}=\dfrac{-v}{u}}}$

hi = Height of image

ho = Height of object

Solution :

$\implies{\sf \dfrac{1}{20}=\dfrac{1}{v}-\left(\dfrac{1}{-80}\right)}$

$\implies{\sf \dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{80} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{80} }$

$\implies{\sf \dfrac{1}{v}=\dfrac{80-(-20)}{20 \times 80}}$

$\implies{\sf \dfrac{1}{v}=\dfrac{100}{1600}}$

$\implies{\sf v =\dfrac{1600}{100}}$

$\implies{\bf v=16\:cm }$

____________________________________

★ Size of image -

$\implies{\sf \dfrac{h_i}{h_o}=\dfrac{-v}{u} }$

$\implies{\sf \dfrac{h_i}{1}=\dfrac{-16}{80} }$

$\implies{\bf h_i =- 0.2\: cm }$

★ Location of image :

• Between F and 2F
• Real
• Inverted
• Small in size ( diminished )

Answer :

Image distance (v) = 16 cm

Height of image (hi) = -0.2 cm

Answer 2

Answer:

Solution

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Verified by Toppr

Correct option is C)

Object distance u=−0.1 m

Focal length f=0.05 m

Needle height h

o

=1 cm

Using mirror formula

v

1

+

u

1

=

f

1

∴

v

1

+

−0.1

1

=

0.05

1

⟹ v=

30

1

cm

Object height h

o

=1 cm

Magnification m=

h

o

h

i

=

u

−v

∴

1

h

i

=

30×(−0.1)

−1

⟹ h

i

=0.33 cm