A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. An object is placed at 20 cm to the left of this lens system. The distance of the image from the lens in cm is

1. Thin Lenses in Contact

If two thin lenses of focal lengths $f_1$ and $f_2$ touch each other, they behave like a single thin lens whose effective focal length $F$ is given by

$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$

For a convex lens $f_1$ is positive; for a concave lens $f_2$ is negative (sign convention: distances measured in the direction of incident light are negative).

2. Lens Formula for a Single (Effective) Lens

After you find $F$, the ordinary thin-lens equation applies:

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

where

• $u$ = object distance (take negative for objects placed to the left of the lens)
• $v$ = image distance (positive to the right of the lens, negative to the left)

3. Logical Chain of Steps

1. Identify focal lengths: $f_1=+30\,\text{cm}$ (convex), $f_2=-20\,\text{cm}$ (concave).
2. Compute $F$ by adding their reciprocals.
3. Plug $u=-20\,\text{cm}$ and the calculated $F$ into the lens formula.
4. Solve for $v$. If you get a negative $v$, the image lies on the same side as the object (virtual).

Imagine Two Magnifying Glasses Stuck Together

• One glass bends light so it converges (comes together).
• The other bends light so it diverges (spreads apart).
• When you glue them so close that they almost touch, they behave like one single special glass.
• To know where the picture (image) will appear when you put a toy (object) in front of this special glass, you just:
  1. Blend the powers of both glasses into one number (called effective focal length).
  2. Use the simple lens formula (a friendly rule that relates where you kept the toy, how strong the glass is, and where the picture shows up).
     That’s all!

1. Effective focal length $F$
   $\frac{1}{F} = \frac{1}{+30\,\text{cm}} + \frac{1}{-20\,\text{cm}}$
   (Work out the arithmetic to find $F$.)

2. Lens formula
   Object distance: $u=-20\,\text{cm}$
   $\frac{1}{F} = \frac{1}{v} - \left( -\frac{1}{20\,\text{cm}} \right)$
   Solve this equation to obtain $v$.

3. Interpretation

   • If $v$ turns out negative, the image lies $|v|$ cm to the left of the lens system (virtual).
   • If $v$ is positive, the image is $v$ cm to the right (real).