Question

An object 4cm high is place at distance of 27cm front of a convex lens of focal length 18cm. Find the position,nature and size of the image formed

Answer 1

$\mathcal{\huge{\underline{\underline{\red{Given:-}}}}}$

• Height of an object is 4cm.

• Distance of 27cm.

• Focal length is 18cm.

$\mathcal{\huge{\underline{\underline{\green{To\:Find:-}}}}}$

▶The position of image

▶Nature of image

▶Size of the image

⭐$\mathcal{\huge{\underline{\underline{\pink{Solution:-}}}}}$

According to the question,

⬛ Finding the position of image :-

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

From given, above

=> $\dfrac{1}{v} = \dfrac{1}{18} + \dfrac{1}{27}$

=> $\dfrac{1}{v} = \dfrac{3}{54} + \dfrac{2}{54}$

=> $\dfrac{1}{v} = \dfrac{5}{54}$

=> ${v} = \dfrac{54}{5}$

=> v = 10.8cm

➡ Hence,the position of image is 10.8cm.

⬛ Nature of image :-

✒ Positive(+ve) sign of v shows that image must be real.

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⬛ Size of the image :-

Formulae :- $m = \dfrac{v}{u} = \dfrac{h'}{h}$

=> $\dfrac{10.8}{27} = \dfrac{h'}{4}$

=> h'= $\frac{43.2}{27}$

=> h'=1.6cm

➡ Hence,the size of the image formed is 1.6 cm.

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