NEET (UG)

Ray Optics & Optical Instruments

Reflection & mirrors, refraction & lenses, total internal reflection, prisms & instruments

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Module 1

Reflection and Refraction

Reflection and Spherical MirrorsTopic 1

Light travels in straight lines and obeys the laws of reflection: the angle of incidence equals the angle of reflection, and the incident ray, reflected ray and normal all lie in one plane. A plane mirror forms a virtual, erect image of the same size, laterally inverted, as far behind the mirror as the object is in front — the everyday case from which the sign conventions are built.

Curved mirrors form the heart of this topic. A concave mirror converges light and a convex mirror diverges it. Both obey the mirror formula $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$, where the focal length is half the radius of curvature, $f = R/2$. Distances are measured from the pole using the Cartesian sign convention, with distances along the incident light taken positive.

The size and orientation of the image are given by the magnification $m = -\dfrac{v}{u} = \dfrac{h'}{h}$. A negative $m$ means an inverted (real) image and a positive $m$ an erect (virtual) image. A concave mirror can give either, depending on where the object is placed, while a convex mirror always gives a small, erect, virtual image — which is why it is used as a rear-view mirror for a wide field of view.

Mastering this topic for NEET is mostly about the sign convention and quick image analysis. The practised approach is to fix the conventions, substitute carefully into the mirror formula, and read off whether the image is real or virtual, erect or inverted, enlarged or diminished — a sequence that turns most mirror problems into a one-line calculation.

Figure — Reflection and Spherical Mirrors
QuantityRelation
Mirror formula$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Focal length$f = R/2$
Magnification$m = -\dfrac{v}{u}$
Worked Examples
1

An object is placed $30\ \text{cm}$ from a concave mirror of focal length $20\ \text{cm}$. Find the image distance.

Show solution

Using $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ with $f = -20$, $u = -30$: $\dfrac{1}{v} = \dfrac{1}{-20} - \dfrac{1}{-30} = -\dfrac{1}{60}$, so $v = -60\ \text{cm}$ — a real, inverted image $60\ \text{cm}$ in front.

2

Why is a convex mirror used as a vehicle rear-view mirror?

Show solution

A convex mirror always forms a small, erect, virtual image and has a wide field of view, so the driver can see a large area behind the vehicle in a compact mirror.

✎ Self-Check — 5 questions0 / 5
Q1.

The focal length of a spherical mirror is related to its radius by:

Explanation: $f = R/2$.
Q2.

The mirror formula is:

Explanation: $1/f = 1/v + 1/u$.
Q3.

A convex mirror always forms an image that is:

Explanation: Convex mirror: always small, erect, virtual.
Q4.

Magnification of a mirror is:

Explanation: $m = -v/u$.
Q5.

A negative magnification indicates an image that is:

Explanation: Negative $m$ means an inverted (real) image.

NEET tip: Fix the sign convention first, then apply $1/f = 1/v + 1/u$ and $m = -v/u$. Convex mirrors always give a small, erect, virtual image; $f = R/2$.

Refraction and Total Internal ReflectionTopic 2

Refraction is the bending of light as it passes between media of different optical densities, caused by a change in its speed. It obeys Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$, where $n = c/v$ is the refractive index. Light bends towards the normal on entering a denser medium and away from it on entering a rarer one — the reason a straw looks bent in a glass of water.

Several familiar effects follow from refraction. A pool looks shallower than it is because of apparent depth, where real depth and apparent depth are related by the refractive index, $n = \dfrac{\text{real depth}}{\text{apparent depth}}$. The twinkling of stars and the appearance of a coin rising in a beaker of water are all consequences of the same bending of light.

When light travels from a denser to a rarer medium, increasing the angle of incidence eventually makes the refracted ray graze the surface. The incidence angle at which this happens is the critical angle $\theta_c$, given by $\sin\theta_c = \dfrac{1}{n}$. Beyond it, light is reflected entirely back into the denser medium — a phenomenon called total internal reflection (TIR).

TIR requires two conditions: light must go from a denser to a rarer medium, and the angle of incidence must exceed the critical angle. It explains the brilliance of diamonds (very high $n$, small $\theta_c$), the formation of mirages, and most importantly the working of optical fibres, where light bounces along the core by repeated total internal reflection — a high-yield NEET application.

Figure — Refraction and Total Internal Reflection
QuantityRelation
Snell's law$n_1\sin\theta_1 = n_2\sin\theta_2$
Refractive index$n = c/v$
Critical angle$\sin\theta_c = 1/n$
Apparent depth$n = \dfrac{\text{real}}{\text{apparent}}$
Worked Examples
1

Find the critical angle for a medium of refractive index $n = 2$.

Show solution

$\sin\theta_c = \dfrac{1}{n} = \dfrac{1}{2}$, so $\theta_c = 30^{\circ}$.

2

A pond appears $3\ \text{m}$ deep. If the refractive index of water is $4/3$, find its real depth.

Show solution

$n = \dfrac{\text{real depth}}{\text{apparent depth}}$, so real depth $= n\times\text{apparent} = \dfrac{4}{3}\times3 = 4\ \text{m}$.

✎ Self-Check — 5 questions0 / 5
Q1.

Snell's law is:

Explanation: $n_1\sin\theta_1 = n_2\sin\theta_2$.
Q2.

The critical angle is given by:

Explanation: $\sin\theta_c = 1/n$.
Q3.

Total internal reflection occurs when light goes from a:

Explanation: Denser to rarer, with angle above the critical angle.
Q4.

Optical fibres work on the principle of:

Explanation: Light is guided by repeated TIR.
Q5.

A medium with a higher refractive index has a critical angle that is:

Explanation: $\sin\theta_c = 1/n$; larger $n$ gives smaller $\theta_c$.

NEET tip: Use $n_1\sin\theta_1 = n_2\sin\theta_2$ and $\sin\theta_c = 1/n$. TIR needs denser-to-rarer travel and incidence above $\theta_c$ — the basis of optical fibres and the sparkle of diamond.

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Module 2

Lenses, Prisms and Optical Instruments

Lenses, the Lens-Maker's Formula and PrismsTopic 3

A lens refracts light at its two curved surfaces. A convex (converging) lens brings parallel rays to a focus, while a concave (diverging) lens spreads them out. The thin-lens behaviour is captured by the lens formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ — note the minus sign, which differs from the mirror formula and is a frequent source of NEET errors.

The focal length itself depends on the lens shape and material through the lens-maker's formula $\dfrac{1}{f} = (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$. A lens has more converging power if it is made of higher-index glass or has more sharply curved surfaces. The power of a lens, $P = \dfrac{1}{f}$ (in dioptres with $f$ in metres), is positive for converging and negative for diverging lenses, and powers simply add for lenses in contact: $P = P_1 + P_2$.

A prism bends light through an angle of deviation $\delta$ that depends on the prism angle $A$, the refractive index, and the angle of incidence. The deviation is least at the angle of minimum deviation $\delta_m$, where the refractive index is given by $n = \dfrac{\sin\left(\dfrac{A + \delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$ — a standard NEET formula.

Because the refractive index varies slightly with colour (highest for violet, lowest for red), a prism bends violet light more than red and splits white light into a spectrum — dispersion. This is the origin of the rainbow, where raindrops act as tiny prisms. Recognising that violet deviates most and red least, and that dispersion arises from index varying with wavelength, answers most prism-colour questions.

Figure — Lenses, the Lens-Maker's Formula and Prisms
QuantityRelation
Lens formula$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Lens-maker's formula$\dfrac{1}{f} = (n-1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$
Power$P = 1/f$ (dioptre)
Prism (min. deviation)$n = \dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}}$
Worked Examples
1

Two thin lenses of powers $+3\ \text{D}$ and $-1\ \text{D}$ are placed in contact. Find the power and focal length of the combination.

Show solution

$P = P_1 + P_2 = 3 + (-1) = 2\ \text{D}$. Focal length $f = \dfrac{1}{P} = \dfrac{1}{2} = 0.5\ \text{m} = 50\ \text{cm}$.

2

In a prism, which colour deviates the most and why?

Show solution

Violet light deviates the most because the refractive index of the prism is highest for violet (shortest wavelength); red, with the lowest index, deviates least.

✎ Self-Check — 5 questions0 / 5
Q1.

The lens formula is:

Explanation: $1/f = 1/v - 1/u$ (note the minus).
Q2.

The power of a lens is:

Explanation: $P = 1/f$ in dioptres.
Q3.

For lenses in contact, the powers:

Explanation: $P = P_1 + P_2 + \dots$.
Q4.

A diverging (concave) lens has a power that is:

Explanation: Concave lens has negative power.
Q5.

Dispersion by a prism occurs because the refractive index depends on:

Explanation: $n$ varies with wavelength (colour).

NEET tip: Lens formula uses a minus: $1/f = 1/v - 1/u$. Power $P = 1/f$ (dioptre, positive for convex) and powers add in contact. Violet deviates most, red least, in a prism.

Optical Instruments and the Human EyeTopic 4

Optical instruments combine lenses to extend what the eye can see. The simple microscope (magnifying glass) is a single convex lens that produces an enlarged virtual image; its magnifying power is $M = 1 + \dfrac{D}{f}$ when the image is at the near point $D$ (about $25\ \text{cm}$). The shorter the focal length, the greater the magnification.

A compound microscope uses two lenses: an objective of short focal length forms a real, magnified image, which the eyepiece then magnifies further. Its total magnification is roughly the product of the two stages, $M \approx \dfrac{L}{f_o}\times\dfrac{D}{f_e}$, so both lenses should have small focal lengths for high magnification — a key design fact NEET tests.

A telescope views distant objects and uses a large focal-length objective with a short focal-length eyepiece. Its magnifying power in normal adjustment is $M = \dfrac{f_o}{f_e}$, the opposite requirement to the microscope. A large objective also gathers more light, giving brighter, sharper images of faint stars — which is why astronomical telescopes have big objective lenses or mirrors.

The human eye focuses light onto the retina, varying its lens power by accommodation. Common defects are corrected with lenses: myopia (short-sightedness, distant objects blurred) is corrected with a concave lens, and hypermetropia (long-sightedness, near objects blurred) with a convex lens. Knowing which lens corrects which defect, and the microscope-versus-telescope focal-length requirements, covers the bulk of NEET instrument questions.

Figure — Optical Instruments and the Human Eye
Instrument / defectKey relation / fix
Simple microscope$M = 1 + D/f$
Compound microscopeshort $f_o$ and $f_e$ (high $M$)
Telescope$M = f_o/f_e$ (large $f_o$)
Myopia / Hypermetropiaconcave / convex lens
Worked Examples
1

A telescope has an objective of focal length $100\ \text{cm}$ and an eyepiece of focal length $5\ \text{cm}$. Find its magnifying power in normal adjustment.

Show solution

$M = \dfrac{f_o}{f_e} = \dfrac{100}{5} = 20$.

2

Which lens corrects myopia (short-sightedness)?

Show solution

A concave (diverging) lens. It diverges the incoming light so that the image of a distant object falls on the retina rather than in front of it.

✎ Self-Check — 5 questions0 / 5
Q1.

The magnifying power of a telescope is:

Explanation: $M = f_o/f_e$.
Q2.

For high magnification, a compound microscope needs:

Explanation: Both lenses should have short focal lengths.
Q3.

Myopia is corrected using a:

Explanation: Concave lens corrects short-sightedness.
Q4.

Hypermetropia is corrected using a:

Explanation: Convex lens corrects long-sightedness.
Q5.

The ability of the eye lens to change its focal length is called:

Explanation: Accommodation lets the eye focus at different distances.

NEET tip: Microscope needs short $f_o$ and $f_e$; telescope needs large $f_o$, small $f_e$ with $M = f_o/f_e$. Myopia to concave lens; hypermetropia to convex lens.

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