A converging lens f=20cm — object at 30cm — find image position and magnification

medium CBSE JEE-MAIN NEET • 3 min read
Tags Optics

Question

An object is placed 30 cm in front of a converging (convex) lens of focal length 20 cm. Find the position of the image and the magnification produced.

Solution — Step by Step

Using the standard sign convention (all distances measured from optical centre; distances in the direction of incident light are positive):

  • Object distance: u=−30u = -30 cm (object is on the left of lens, so negative)
  • Focal length: f=+20f = +20 cm (converging lens, so positive)
  • Image distance: v=?v = ?

The lens formula is:

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

Substituting:

1v−1−30=120\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} 1v+130=120\frac{1}{v} + \frac{1}{30} = \frac{1}{20} 1v=120−130\frac{1}{v} = \frac{1}{20} - \frac{1}{30}

Finding a common denominator (LCM of 20 and 30 is 60):

1v=360−260=160\frac{1}{v} = \frac{3}{60} - \frac{2}{60} = \frac{1}{60} v=+60 cm\boxed{v = +60\text{ cm}}

The positive sign means the image is formed on the opposite side of the lens from the object — this is a real image.

Linear magnification:

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

Interpretation: The magnitude 2 means the image is twice the size of the object. The negative sign means the image is inverted (real and inverted).

Why This Works

The lens formula 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} is derived from Snell’s law applied to paraxial rays through a thin lens. Each surface of the lens refracts light, and the combined effect is captured in this single formula.

The sign of vv tells us the image type:

  • v>0v > 0: real image (on the transmission side, can be captured on screen)
  • v<0v < 0: virtual image (on the same side as the object, cannot be on screen)

Here, the object is between ff and 2f2f (20 cm and 40 cm), at 30 cm. For a convex lens, an object between ff and 2f2f produces a real, inverted, magnified image beyond 2f2f. Our answer of 60 cm confirms this — 60 cm is beyond 40 cm (2f2f).

Alternative Method — Using Cartesian Form

Some textbooks write the lens formula as 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} and others as 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}. These are identical. The key is consistent application of the sign convention. Using u=−30u = -30:

1f=1v−1u  ⟹  120=1v+130\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \implies \frac{1}{20} = \frac{1}{v} + \frac{1}{30}

Same calculation, same answer.

Common Mistake

Using u=+30u = +30 (forgetting the negative sign). In the standard sign convention, object distance uu is always negative for a real object (because the object is on the same side as incoming light, which is the negative direction). Using u=+30u = +30 gives 1v=120−130=160\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}… accidentally getting the right numerical answer but for the wrong reason. When uu is substituted with wrong sign into other problems, you get completely wrong results.

Quick check for convex lens: when object is at 2f=402f = 40 cm, image is also at 2f=402f = 40 cm and magnification = 1 (same size). Our object at 30 cm is between ff and 2f2f, so image should be beyond 2f2f and magnified. Our answer: v=60v = 60 cm (beyond 40 cm) and ∣m∣=2|m| = 2 (magnified). This mental check confirms we’re on the right track.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

A Myopic Person Needs -2 D Lens — What is Their Far Point?

hard

Dispersion of light — prism, rainbow formation, chromatic aberration

medium

Lens and mirror formula problems — step by step approach with sign convention

medium

Mirror Formula 1/v + 1/u = 1/f — Sign Convention and Example

easy

Object 30 cm in Front of Concave Mirror, f = 15 cm — Find Image

medium