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Ray Optics: concepts, formulas & solved PYQ-style examples

Ray optics is a high-weightage chapter in NEET physics and appears regularly in JEE. Nearly every numerical comes down to the mirror or lens formula with the correct sign convention.

Measure all distances from the pole or optical centre; distances in the direction of incident light are positive.

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Key concepts

Sign convention

Cartesian convention: object distance u is negative for real objects.

Mirror formula

Concave mirrors form virtual, magnified images when the object is inside f.

Lens formula

Convex lenses form real, inverted images beyond f; lens maker’s formula gives f.

Refraction and TIR

Snell’s law, critical angle, optical fibres and apparent depth.

Prism

Angle of deviation and the minimum-deviation condition.

Formula sheet

Mirror formula

1v+1u=1f,m=−vu\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f},\quad m = -\dfrac{v}{u}

Lens formula

1v−1u=1f,m=vu\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f},\quad m = \dfrac{v}{u}

Lens maker’s formula

1f=(μ−1)(1R1−1R2)\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)

Snell’s law, critical angle

μ1sin⁡i=μ2sin⁡r,sin⁡C=1μ\mu_1\sin i = \mu_2\sin r,\quad \sin C = \dfrac{1}{\mu}

Prism at minimum deviation

μ=sin⁡A+δm2sin⁡A2\mu = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}

What PYQs keep asking

Patterns that repeat across previous-year JEE and NEET papers for this chapter.

  • Image position and nature for an object placed inside or beyond the focus.
  • Lens maker’s formula, including a lens immersed in a liquid.
  • Critical angle and total internal reflection conditions.
  • Minimum deviation through a prism.

Solved examples

Example 1

An object is placed 10 cm in front of a concave mirror of focal length 15 cm. Find the position and nature of the image.

  1. 1Sign convention: u=−10u = -10 cm, f=−15f = -15 cm.
  2. 21v=1f−1u=−115+110=130⇒v=+30\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{1}{30} \Rightarrow v = +30 cm.
  3. 3m=−vu=−30−10=+3m = -\dfrac{v}{u} = -\dfrac{30}{-10} = +3.
Answer: Image 30 cm behind the mirror — virtual, erect, 3× magnified
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Example 2

An object is 30 cm from a convex lens of focal length 20 cm. Find the image distance and magnification.

  1. 1u=−30u = -30 cm, f=+20f = +20 cm.
  2. 21v=1f+1u=120−130=160⇒v=+60\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{1}{60} \Rightarrow v = +60 cm.
  3. 3m=vu=60−30=−2m = \dfrac{v}{u} = \dfrac{60}{-30} = -2.
Answer: v = 60 cm on the other side — real, inverted, 2× magnified
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Common traps

  • Dropping the minus sign of u or f for mirrors.
  • Using the mirror formula (plus sign) for lenses (minus sign).
  • Forgetting that a lens’s focal length changes when it is placed in a liquid.

Practise with

Modern PhysicsKinematics & Projectile MotionCurrent ElectricityLens ray diagram simulationRefraction and total internal reflection simulation295 Ray Optics practice questionsVideo solutions

FAQ

When does a concave mirror form a virtual image?

When the object is between the pole and the focus; the image is then virtual, erect and magnified.

Why is the lens formula different from the mirror formula?

Both use the same sign convention, but light passes through a lens, which changes the relation to 1/v − 1/u = 1/f.

Is ray optics important for NEET?

Yes, it is a high-weightage NEET physics chapter, with frequent mirror, lens and prism numericals.

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