Lens formula 1/v - 1/u = 1/f for concave and convex lens numerical

Question

An object is placed 30 cm from a convex lens of focal length 20 cm. Find the position, nature, and magnification of the image. Repeat for a concave lens of focal length 15 cm with the object at 30 cm.

(CBSE 2024, similar pattern)

Solution — Step by Step

Part 1: Convex Lens

Using the New Cartesian Sign Convention: object is always on the left, so $u = -30$ cm. For a convex lens, $f = +20$ cm (positive focal length).

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

\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}

\frac{1}{v} + \frac{1}{30} = \frac{1}{20}

\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}

v = +60 \text{ cm}

Positive $v$ means the image is on the right side — it is real and inverted.

m = \frac{v}{u} = \frac{60}{-30} = -2

The image is 2 times the size of the object (magnified) and inverted (negative sign).

Part 2: Concave Lens

For a concave lens: $f = -15$ cm, $u = -30$ cm.

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-15} + \frac{1}{-30} = \frac{-2 - 1}{30} = \frac{-3}{30} = \frac{-1}{10}

v = -10 \text{ cm}

Negative $v$ means the image is on the same side as the object — it is virtual, erect, and diminished.

m = \frac{v}{u} = \frac{-10}{-30} = +\frac{1}{3}

The image is 1/3 the size of the object and erect (positive sign).

Why This Works

The lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ is derived from geometry using similar triangles formed by rays passing through the optical centre and focus. The sign convention ensures the formula works for both convex and concave lenses without modification.

Key pattern: a concave lens ALWAYS forms a virtual, erect, diminished image regardless of object position. A convex lens forms different types of images depending on where the object is placed relative to $F$ and $2F$ .

Alternative Method

You can also use the magnification formula $m = \frac{f}{f - u}$ (derived from combining lens and magnification formulas). For the convex lens: $m = \frac{20}{20 - (-30)} = \frac{20}{50} = -2$ … wait, that gives the wrong sign. The correct derived formula is $m = \frac{f}{f + u}$ when using the convention properly. Stick with $m = v/u$ to avoid sign errors.

Quick check: for a convex lens, if the object is beyond $2F$ ( $u > 2f$ ), the image is between $F$ and $2F$ (diminished). If the object is between $F$ and $2F$ , the image is beyond $2F$ (magnified). Here, $u = 30$ cm and $2f = 40$ cm, so object is between $F$ and $2F$ — image should be magnified. Our answer ( $m = -2$ ) confirms this.

Common Mistake

The number one error: forgetting to apply the sign convention and using all positive values. The object distance $u$ is ALWAYS negative (object on left). For a concave lens, $f$ is ALWAYS negative. If you skip sign convention, you will get wrong image positions and wrong nature of image. Write down the signs explicitly before substituting into the formula.