Question

A biconvex lens has a radius of curvature of magnitude $20\, cm$. Which one of the following options describe best the image formed of an object of height $2 \,cm$ placed $30\, cm$ from the lens?

• A. Virtual, upright, height = 1 cm
• B. Virtual, upright, height = 0.5 cm
• C. Real, inverted, height = 4 cm
• D. Real, inverted, height = 1 cm

Answer 1

The Correct Option is C

Focal length of the lens
$\frac{1}{ f }=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)=\frac{1}{20}$
$f =20 \,cm$
From lens formula
$\frac{1}{ v }-\frac{1}{ u }=\frac{1}{ f }$
$\frac{1}{ v }-\frac{1}{-30}=\frac{1}{20}$
$v =60\, cm$
$\frac{ I }{ O }= m =\frac{ v }{ u }=\frac{60}{-30}=- I$
$I =-2(0)=-2 \times 2=-4 cm$
so image will be real inverted and of size $4\, cm$.

Answer 2

Given:

 

The radii of curvature, R₁=R₂=20 cm

Object distance, u=30 cm (the object is placed on the left side, so u is negative)

Height h_o = 2 cm

Let us consider the light incident from the air on a glass lens, so 

μ = \(\frac{3}{2}\). Also, for a biconvex lens, R₁ is positive and R₂ is negative.

 

Therefore, in the case of the biconvex lens,

 

\(\frac{1}{f}\) = (μ−1)(\(\frac{1}{R_1}-\frac{1}{R_2}\))

⇒\(\frac{1}{f}\) =(\(\frac{3}{2-1}\))[\(\frac{1}{20}-(\frac{-1}{20})\)]

⇒\(\frac{1}{f}\) = (\(3-\frac{2}{2}\))(\(\frac{1}{20}+\frac{1}{20}\))

⇒\(\frac{1}{f}\) = \(\frac{1}{2}\times(\frac{2}{20})\)

∴\(\frac{1}{f}\) =\(\frac{1}{20}\)

So,

f = 20cm.

Now, calculate image distance by substituting the values of focal length and object distance in the thin lens formula:

\(\frac{1}{f}\) =\(\frac{1}{v}\)−\(\frac{1}{u}\)

Applying the values f = 20 and u=30 we get,

V = 60

⇒ So, the image distance is 60 cm from the lens and the image is formed on its right side. This is the reason why the image looks real in nature.

Now, We can measure the magnification of the object by the lens,

m=−\(\frac{H_i}{h_o}\) = −\(\frac{v}{u}\)

\(h_i\) = \(\frac{v}{u}\times h_o\)

⇒ \(h_i\) = \(\frac{60}{-30}\)×2

∴ \(h_i\)=−4 cm

The negative sign of the image’s height signifies that the image is inverted.

Therefore, option C is the correct answer.

 

Answer 3

In general, we have to assume 

μ=1.5

So, f=20cm

\(\frac{1}{f}\) = \(\frac{1}{v}+\frac{1}{u}\) ⇒ \(\frac{1}{20}\) = \(\frac{1}{v}+\frac{1}{30}\)

\(\frac{1}{v}\)= \(\frac{1}{20}-\frac{1}{30}\)=\(\frac{10}{600}\)

V =60cm

\(\frac{h_i}{h_0}\) = 2

h_i = 2×|h₀|  ⇒  h_i = 4cm

Here, the image is real, inverted, magnified field and the height of the image is 

4cm.

So, the correct option will be (C).