This question involves calculating the ratio of magnifications for a lens at two different object positions. Let's break it down step by step.

Key Concepts:

A biconvex lens is a converging lens. The lens formula relates the object distance (u), image distance (v), and focal length (f):

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

Magnification (m) is defined as the ratio of the height of the image to the height of the object and is also given by:

$m=\frac{v}{u}$

Note: According to the sign convention, for a convex lens, the focal length (f) is positive, and the object distance (u) is negative. Therefore, we will use $f=+20\mathrm{cm}$, $u_1=-25\mathrm{cm}$, and $u_2=-50\mathrm{cm}$.

Step-by-Step Solution:

Step 1: Find the image distance (v₁) when u = -25 cm.

Using the lens formula: $\frac{1}{v_1}=\frac{1}{f}+\frac{1}{u_1}$

Substitute the values: $\frac{1}{v_1}=\frac{1}{20}+\frac{1}{-25}=\frac{1}{20}-\frac{1}{25}$

Find a common denominator (100): $\frac{1}{v_1}=\frac{5}{100}-\frac{4}{100}=\frac{1}{100}$

Therefore, $v_1=100\mathrm{cm}$.

Now, calculate the magnification m₂₅: $m_{25}=\frac{v_1}{u_1}=\frac{100}{-25}=-4$

Step 2: Find the image distance (v₂) when u = -50 cm.

Using the lens formula again: $\frac{1}{v_2}=\frac{1}{f}+\frac{1}{u_2}$

Substitute the values: $\frac{1}{v_2}=\frac{1}{20}+\frac{1}{-50}=\frac{1}{20}-\frac{1}{50}$

Find a common denominator (100): $\frac{1}{v_2}=\frac{5}{100}-\frac{2}{100}=\frac{3}{100}$

Therefore, $v_2=\frac{100}{3}\mathrm{cm}$.

Now, calculate the magnification m₅₀: $m_{50}=\frac{v_2}{u_2}=\frac{\frac{100}{3}}{-50}=-\frac{100}{3\times 50}=-\frac{2}{3}$

Step 3: Find the ratio m₂₅ / m₅₀.

We have $m_{25}=-4$ and $m_{50}=-\frac{2}{3}$.

The ratio is: $\frac{m_{25}}{m_{50}}=\frac{-4}{-\frac{2}{3}}=4\times \frac{3}{2}=6$

Final Answer: The ratio $\frac{m_{25}}{m_{50}}$ is $6$.

Related Topics & Formulae:

Lens Formula: The fundamental equation for thin lenses is $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$, where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. Remember to use the sign convention (u is negative for real objects).

Linear Magnification: It is given by $m=\frac{v}{u}=\frac{h}{i}ho$, where $h_i$ is the image height and $h_o$ is the object height. A negative value indicates an inverted image.

Converging Lens Behavior: A biconvex lens is converging. For a real object placed beyond the focal point (u > f), it forms a real and inverted image. As the object moves away from the lens, the image moves closer to the focal point and becomes smaller, which is why the magnification changes.