Explanation

We solve the problem using strict Cartesian sign convention, which is mandatory for JEE Main, JEE Advanced and IIT JEE.

Case 1: Lenses in contact

Equivalent focal length:

$$ \frac{1}{f}=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20} \Rightarrow f=-20\text{ cm} $$

Object distance:

$$ u=-10\text{ cm} $$

Lens formula:

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

$$ \frac{1}{v}+\frac{1}{10}=-\frac{1}{20} \Rightarrow \frac{1}{v}=-\frac{3}{20} $$

$$ v=-\frac{20}{3}\text{ cm} $$

Magnification:

$$ m_1=\frac{v}{u}=\frac{-20/3}{-10}=\frac{2}{3} $$

Case 2: Lenses separated by 1 cm

For convex lens:

$$ \frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1} $$

$$ \frac{1}{v_1}+\frac{1}{10}=\frac{1}{5} \Rightarrow v_1=10\text{ cm} $$

The concave lens is 1 cm to the right, hence the image lies 9 cm to its right:

$$ u_2=+9\text{ cm} $$

For concave lens:

$$ \frac{1}{v_2}-\frac{1}{u_2}=\frac{1}{f_2} $$

$$ \frac{1}{v_2}-\frac{1}{9}=-\frac{1}{4} \Rightarrow \frac{1}{v_2}=-\frac{5}{36} $$

$$ v_2=-\frac{36}{5}\text{ cm} $$

Magnifications:

$$ m_{\text{convex}}=\frac{10}{-10}=-1 $$

$$ m_{\text{concave}}=\frac{-36/5}{9}=-\frac{4}{5} $$

Total magnification:

$$ m_2=(-1)\times\left(-\frac{4}{5}\right)=\frac{4}{5} $$

Finally,

$$ \left|\frac{m_1}{m_2}\right|=\frac{2/3}{4/5}=\frac{5}{6} $$

Hence, the physically correct answer is:

$\dfrac{5}{6}$, which is not present in the given options.

Related JEE Main Questions