Question

A 2.0 -cm-high object is placed 12cm from a convex lens perpendicular to its principal axis.The lens forms a real image whose size is 1.5cm .Find the power of the lens.

Answer With Full Explanation

❌ Copied Answer will directly Reported on the Spot ❌

[ Hint :- Answer is 175/9 D ]
​

Answer 1

Explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• Height of the object = 2cm

• Distance of object = 12cm

• Height of image = 1.5cm

• Type of lens - Convex lens

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The power of the lens.

$\bf{\underline{\underline\green{Solution:-}}}$

To find the power of the lens, first we need to find the focal length.

In convex lens

• $\rm Height \: of \: object \: (h_{o} )= + ve$

• $\rm Height \: of \: image \: (h_{i} )= - ve$

• $\rm Distance \: of \: object \: (u) = - ve$

• $\rm Distance \: of \: image \:(v) = + ve$

As we know that:-

$\boxed{\rm \leadsto Magnification = \frac{ h_{i} }h_{o} = \frac{v}{u} }$

Here:-

• $\rm h_{i} = - 1.5cm$

• $\rm h_{o} = 2cm$

•$\rm u = -12cm$

• $\rm v = ?$

Substituting the values:-

$\implies \rm m= \dfrac{ - 1.5}{2} = \dfrac{v}{ - 12}$

$\implies\rm v = \dfrac{ - 1.5 \times 12}{2}$

$\implies\rm v =9cm$

Let us find the focal length

By using lens formula

$\boxed{\rm \leadsto \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }$

$\implies {\rm \dfrac{1}{f} = \dfrac{1}{9} - \dfrac{1}{( - 12)} }$

$\implies {\rm \dfrac{1}{f} = \dfrac{1}{9} + \dfrac{1}{12} }$

$\implies {\rm \dfrac{1}{f} = \dfrac{4 + 3}{36} }$

$\implies {\rm \dfrac{1}{f} = \dfrac{7}{36} }$

$\implies {\rm f = \dfrac{36}{7} cm }$

Converting 36/7cm to metres

$\implies {\rm f = \dfrac{36}{700} m }$

$\boxed{\leadsto {\rm Power = \dfrac{1}{focal \: length(in \: metres)} }}$

${\implies {\rm P= \dfrac{1}{ \frac{36}{700} } }}$

$\implies {\rm P= \dfrac{700}{36} = \dfrac{175}{9} }$

$\implies {\rm P= 19.44 \: diaoptre }$

$\large\underline{\boxed{ \purple{\therefore \rm Power \: of \: lens \: = 19.44 \: D}} }$

Answer 2

Answer: Power of given lens is 175/9 D

Explanation:

Power of lens = 1/f

Given :

Object distance = -12 cm = 0.12 m

Object height ⇒  $h_o$  = 2 cm = 0.02 m

Real image is inverted. Hence, image height ⇒ $h_i$ = -1.5 cm = -0.015 m

Lens formula relates image distance, object distance and focal length of lens.

According to lens formula:

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

v ⇒ image distance

u ⇒ object distance

f ⇒ focal length of lens

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

Multiplying by u on both sides,

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

Magnification of lens ⇒ $m = \frac{v}{u} = \frac{h_i}{h_o}$

⇒  $\frac{1}{m} - 1 = \frac{u}{f}$

⇒  $\frac{h_o}{h_i} - 1 = \frac{u}{f}$

⇒  $\frac{0.02}{-0.015} - 1 = \frac{-0.12}{f}$

⇒  $\frac{-4}{3} - 1 = \frac{-12}{100f}$

⇒  $\frac{-7}{3} = \frac{-3}{25f}$

⇒  $\frac{7}{3} = \frac{3}{25f}$

⇒  $\frac{7}{3}*\frac{25}{3} = \frac{1}{f}$

⇒  $\frac{175}{9} = \frac{1}{f}$

Power of lens  = 175/9 D

#SPJ2