Question

A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20cm the distance of the object from the lens is 30 cm determine the size of the image formed *

+60cm
+10cm
-60cm
-10cm

Answer 1

Given :-

Height of object= 5cm

Focal length (f) = +20

Object distance (u)= -30

To find :-

Size of image formed

• First find the distance of image (v)

By lens formula

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

$\longrightarrow\sf \dfrac{1}{20}= \dfrac{1}{v}-\dfrac{1}{-30}\\ \\ \longrightarrow\sf \dfrac{1}{20}+\dfrac{1}{-30}= \dfrac{1}{v}\\ \\ \longrightarrow\sf \dfrac{1}{20}-\dfrac{1}{30}= \dfrac{1}{v}\\ \\ \longrightarrow\sf \dfrac{3-2}{60}=\dfrac{1}{v}\\ \\ \longrightarrow\sf \dfrac{1}{60}=\dfrac{1}{v}\\ \\ \longrightarrow\sf v= {\boxed{\sf\ 60\ cm}}$

Now we have to find the size of image

According to formula

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

$\longrightarrow\sf \dfrac{v}{u}= \dfrac{h_i}{h_o}\\ \\ \longrightarrow\sf \cancel{\dfrac{60}{-30}}= \dfrac{h_i}{5}\\ \\ \longrightarrow\sf -2\times 5= h_i\\ \\ \longrightarrow\sf -10= h_i\\ \\ \longrightarrow\sf h_i = {\boxed{\sf\ -10cm}}$

• Correct option is 4

$\underline{\boxed{\sf\ \ size \ of \ image = -10cm}}$

Answer 2

$\mathcal{\huge{\underline{\underline{\red{QUESTION ?}}}}}$

✒ A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20cm the distance of the object from the lens is 30 cm determine the size of the image formed ?

• +60cm
• +60cm+10cm
• +60cm+10cm-60cm
• +60cm+10cm-60cm-10cm

$\mathfrak{\huge{\underline{\underline{\green{ANSWER:-}}}}}$

☑ The option D {-10 cm} is correct answer.

$\mathbb{\huge{\underline{\underline{\blue{SOLUTION:-}}}}}$

⭐ Given :-

Height of object (h) = 5cm

Focal length (f) = +20 cm

Object distance (u)= -30 cm

⭐ To find :-

The size of image formed by the lens.

✅ Calculation :-

The distance of image v by :-

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

$=> \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{ - 30}$

$= > \dfrac{1}{20} + \dfrac{1}{ - 30} = \dfrac{1}{v}$

$= > \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{1}{v}$

$=> \dfrac{3 - 2}{60} = \dfrac{1}{v}$

$=> \dfrac{1}{60} = \dfrac{1}{v}$

$= > v \: = 60cm$

But, size of image is left to be found

$m = \dfrac{v}{u} = \dfrac{hi}{ho}$

$=> \dfrac{ - 60}{30} = \dfrac{hi}{5}$

$=> - 2 \times 5 = hi$

$= > - 2 \times 5 = hi$

=> hi = -10 cm

✔ So , The height of image is -10 cm.

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