Question

For a given less,the magnification was found to be twice as large as when the object was 0.15 m distant from it as when the distance was 0.2 m. The focal length of the lens is 
1. 0.15 m
2. 0.20 m
3. 0.10 m 
4. 0.05 m

Answer 1

Correct Answer - Option 3 : 0.10 m 
CONCEPT:

Linear magnification (m):

• It is defined as the ratio of the height of the image (hi) to the height of the object (ho).

\(m = \frac{{{h_i}}}{{{h_o}}}\)

• The ratio of image distance to the object distance is called linear magnification.

\(m = \frac{{image\;distance\;\left( v \right)}}{{object\;distance\;\left( u \right)}} = - \frac{v}{u}\)

• A positive value of magnification means virtual an erect image.
• A negative value of magnification means a real and inverted image.

CALCULATION:

Given, U₁ = 0.15 M, U₂ = 0.2 M,  m₁ = 2 m₂

• The given condition can be written as

\(⇒ \frac{V_{1}}{U_{1}} =2 \frac{V_{2}}{U_{2}}\)

Substituting the values of U₁ & U₂ in the above equation and rewriting the equation for V₁ and V₂

\(⇒ {V_{1}} =2 V_{2} \frac{u_{1}}{U_{2}}\)

\(⇒ {V_{1}} =2 V_{2} \frac{0.15}{0.2}\)

\(⇒ {V_{1}} = \frac{3}{2} V_{2} \)     ------- (1)

• The focal length must remain the same in both times when the object is placed in 0.15 m, and 0.20 m then

\(⇒ \frac{1}{V_{1}} -\frac{1}{U_{1}} = \frac{1}{V_{2}} - \frac{1}{U_{2}}\)

Substituting the value of V₁ from equation 1 to the above equation

\(⇒ \frac{2}{ {3}{} V_{2}} -\frac{1}{V_{2}} = \frac{1}{U_{1}} - \frac{1}{U_{2}}\)

\(⇒ -\frac{1}{3 V_{2}} =\frac{U_{2} -U_{1}}{U_{1}U_{2}}\)

Taking the reciprocal and finding the value of V₂ from the above equation

⇒ V₂ = 20 cm

• The value V₁ is given by

\(⇒ {V_{1}} = \frac{3}{2}\times 20 = 30 cm\)

Then the value of focal length is given by

\(⇒ \frac{1}{f} = \frac{1}{V_{1}} -\frac{1}{U_{1}} \)

\(⇒ \frac{1}{f} = \frac{1}{0.30} +\frac{1}{0.15} \)

\(⇒ {f} = \frac{0.450}{0.45} = 0.10 m\)

• Hence, option 3 is the answer