NCERT Solutions Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Why the Cartesian sign convention matters. Many students get a confused mix of positives and negatives because they swap conventions mid-problem. Lock it in once: (i) light travels from object to mirror along the negative direction (so u is negative for a real object in front of the mirror); (ii) for a concave mirror, both f and R are negative; (iii) for a convex mirror, both are positive. If your final v comes out negative, the image is real and in front; positive means virtual and behind.

Graphical ray-tracing check. Draw three principal rays from the candle tip: (i) a ray parallel to the axis that reflects through the focus; (ii) a ray through the centre of curvature that returns along itself; (iii) a ray through the focus that reflects parallel to the axis. They converge 54 cm in front of the mirror, below the axis, twice as tall - exactly matching our algebra.

Common mistake. Writing f = +18 cm for a concave mirror because "focal length is positive." The sign depends on the convention; in Cartesian convention used by NCERT, concave-mirror focal length is negative. If you forget, you can still get the right magnitude - but the sign analysis (real vs virtual) will be wrong.

Real-world link - shaving and make-up mirrors. Concave mirrors with object inside focus produce an erect, enlarged virtual image - that's why they're used for shaving and applying make-up. Beyond focus, they form a real, inverted image - the principle behind reflecting telescopes and dental mirrors used to "image" a tooth onto the dentist's eye.

JEE/NEET twist. Examiners often phrase the question as "the candle is moved between centre of curvature and focus": there 2f < |u| < f (i.e., 36 > |u| > 18), and the image forms beyond 2f, real, inverted, magnified - exactly the configuration here at |u|=27 cm. Memorise the six standard positions for a concave mirror and you'll never need the formula in MCQs.

Why convex mirrors are everywhere in cars and shops. Two virtues: (i) they always form an erect, diminished virtual image - so a wide field-of-view is compressed into a small mirror, perfect for blind-spot detection; (ii) the image distance is bounded (0 < v < f), so distant objects don't "jump out" at you - they smoothly compress toward the focus.

The "objects in mirror are closer than they appear" warning. A convex mirror makes objects look smaller, which our brain interprets as farther away. So the actual distance to the car behind you is less than your gut estimate - hence the legal warning on every passenger-side mirror in the US/EU.

Alternative - graphical method. For a convex mirror, draw: (i) a ray parallel to the axis that reflects as if coming from the (virtual) focus behind the mirror; (ii) a ray heading toward the centre of curvature that returns along itself. These two reflected rays diverge - extend them backward (dashed lines) until they meet behind the mirror; that's where the virtual image forms.

Common mistake. Writing f = -15 cm for a convex mirror. In the Cartesian convention, convex-mirror f is positive; the focus is behind the mirror, in the +ve direction. Equally common: forgetting to take the absolute value when reporting magnification - m here is a fraction 5/9, not negative.

JEE/NEET tip. For a convex mirror, since f>0 and u<0, the formula 1/v = 1/f - 1/u always gives 1/v > 0, i.e., v is always positive. This algebraically proves: a convex mirror always forms a virtual image for a real object - no exceptions.

Where the apparent-depth formula comes from. Apply Snell's law to a near-vertical ray leaving the bottom of the tank. For small angles, sinθθ. A ray from the needle at real depth d hitting the surface at horizontal distance x has tan₁ = x/d inside, tan₂ = x/d_app outside. Snell: μ sin₁ = sin₂. Small-angle: μ· x/d = x/d_app, giving μ = d/d_app. This is only valid for paraxial viewing (looking straight down) - at oblique angles, apparent depth is more complicated.

Why the microscope moves DOWN, not up. The new liquid is denser (μ'=1.63 > 1.33), so it bends light more - the needle's image appears even closer to the surface than in water. To re-focus on the new, shallower apparent depth, the microscope's objective must descend (toward the water surface).

Real-world link - fishing. Spearing a fish underwater is hard because the fish appears higher than it actually is. Experienced fishers aim below the apparent position. Conversely, looking up from underwater, the world above looks compressed into a circle (Snell's window) due to total internal reflection at the boundary.

Common mistake. Reporting the shift as 12.5 - 7.67 (i.e., from real depth to new apparent depth). But the microscope was originally focused at the water's apparent depth 9.4 cm, not the real depth - so the relevant difference is 9.4 - 7.67, not 12.5 - 7.67.

Why chain through air. Refractive indices are defined relative to vacuum or, with ∼ 0.03% error, to air. Whenever you go from medium 1 to medium 2, the relevant ratio is ₁₂ = ₂/₁. Memorise: μ for water ≈ 1.33, glass ≈ 1.5, diamond ≈ 2.42. The "denser" medium larger μ bends light toward the normal.

Alternative - direct Snell's chain. You can also write _w sin 45° = _g sin_g directly, then plug in textbook values: 1.33 × 0.707 = 1.5 sin_g, giving sin_g = 0.627, _g ≈ 38.8° - same answer to one decimal. The minor discrepancy with our 38.5° comes from using figure-derived μ's rather than book values.

Real-world link - fibre optics. Glass-to-water refraction is the same physics that lets glass fibres trap light. As long as the angle exceeds the critical angle, total internal reflection kicks in and the ray bounces along the fibre with negligible loss. We'll explore this in Q 9.17.

Common mistake. Confusing "refractive index of glass with respect to water" with the ratio the other way. The rule is: _ab = v_a/v_b = _b/_a (where the LHS uses "with respect to" and the RHS uses absolute indices). Pure memorisation often goes wrong - always derive from Snell's law.

Snell's window. If you're swimming underwater and look up, you see the entire world above compressed into a circular patch overhead - this is the same effect. Outside the patch, the water surface acts like a mirror, reflecting the bottom of the pool. Fish use this all the time to spot prey above and predators on the surface.

Compact formula. Combine the steps: r = htan_c, and tan_c = sin_c/√1-sin²_c = 1/μ/√1-1/μ² = 1/√μ²-1. So

r =

h

√μ²-1

, A =

π h²

μ²-1

.

Plug in: A = π(0.8)²/1.33²-1 = π(0.64)/(0.7689) = 2.61 m². Cleaner.

Why deeper bulbs light bigger patches. The area scales as h² - doubling the depth quadruples the lit patch. But the intensity (power per area) drops correspondingly - Newton's inverse-square law of light in disguise.

Common mistake. Computing _c from sin_c = μ/1 = 1.33, which gives an impossible "sin > 1." The convention is: critical angle is measured in the denser medium when going to the rarer one. Always set up sin_c = n_rarer/n_denser - less than 1.

Why immersion in water reduces deviation. Deviation depends on the ratio of refractive indices at the interface. In air, the ratio is 1.53/1 = 1.53; in water it's 1.53/1.33 ≈ 1.15. Smaller contrast = less bending. If you could match the surrounding medium's index exactly to the glass _medium = _glass, the prism would become "invisible" - no refraction at all. This is how immersion oil works in microscopy.

Alternative - small-angle limit. For thin prisms (small A), D_m ≈ μ - 1A. In air: D_m = (1.53-1)× 60° = 31.8° - close to 40° but the formula is only approximate. In water: D_m' = (1.152-1)× 60° = 9.1° - close to our exact 10°. The thin-prism formula is great for quick estimates.

Real-world link - prism spectroscopes. The dispersive power of a prism (separating colours) comes from the wavelength-dependence of μ. When you immerse a prism in water, dispersion drops too - that's why immersion liquids are used to "tune" the working bandwidth of optical instruments.

Common mistake. Using _g (1.53) directly in the formula for water-immersed case. The formula always uses the relative index between prism and surroundings, not the absolute index.

Why the lens needs a bigger R than f. Notice R = 22 cm is larger than f = 20 cm. In a low-index glass μ = 1.55, the bending is mild, so to achieve a short focal length we need a fairly flat (large-R) surface combination. If you use a higher-index material e.g. dense flint, μ = 1.7, R = 2(0.7)(20) = 28 cm - wait, that's larger. Let me re-examine: actually R = 2μ-1f increases with μ, so higher-index glass needs flatter surfaces to give the same focal length. That's why high-index camera lenses can be thinner.

Alternative method - thin-lens equation as a check. Place an object at u = -40 cm (twice the focal length). For an ideal thin lens with f = 20, we expect v = +40 cm and m = -1. Plug into 1/v - 1/u = 1/f: 1/v = 1/20 - 1/40 = 1/40, so v = +40 cm. Check.

Common mistake. Forgetting the sign of R₂ (writing both positive). That gives 1/f = μ-1(0) = 0 - infinite focal length, which is nonsense. The Cartesian convention is the only way to keep the signs straight.

Real-world link - Aspheric lenses. In premium cameras and eyeglasses, the spherical surfaces (constant R) are replaced by aspheric surfaces (varying curvature) to reduce spherical aberration. Lensmaker's formula no longer directly applies - one has to solve the full wave equation numerically.

Why a concave lens shifts the focus farther. Concave lenses diverge; they "push out" the convergence point. In (b), the original beam was converging to a point 12 cm past the lens, but the concave lens makes the rays diverge a little, delaying convergence to 48 cm. A converging beam can never be made to diverge entirely by a single thin diverging lens unless the diverging power exceeds the original convergence.

Why a convex lens brings it closer. A convex lens adds convergence to the already-convergent beam, pulling the focus closer to the lens - here, from 12 cm down to 7.5 cm.

Alternative method - power addition. Define the original beam's "power" as that of a converging system focusing at 12 cm: P₁ = 1/0.12 = 8.33 D. Add the lens power. Convex f=20 cm: P₂ = 5 D, total 13.33 D, new focus at 1/13.33 = 0.075 m = 7.5 cm. Concave f=-16 cm: P₂ = -6.25 D, total 2.083 D, new focus at 0.48 m = 48 cm. Power addition is a fast shortcut for thin-lens combinations.

Common mistake. Writing u = -12 cm by reflex. Virtual object - light is converging toward the lens - gives positive u. The sign convention treats the object location relative to the direction of incident light, not "in front" or "behind."

The "iron rule" for concave lenses. For ANY real object placed in front of a concave lens, the image is always: virtual, erect, diminished, and located between the optical centre and the focus on the same side as the object. You can prove this algebraically - f<0 and u<0 always give 1/v < 0, and |v| < |f| for any finite |u|.

Real-world link - diverging eyeglasses. Myopia (short-sightedness) is corrected with diverging (concave) lenses. The lens forms a virtual, diminished image of distant objects at a distance the myopic eye can focus - typically a few metres in front. The eye then sees this nearby virtual image rather than the actual distant object.

Alternative - ray diagram. Draw: (i) parallel ray refracts as if diverging from focus on object side; (ii) ray through optical centre passes undeviated. Extend the diverging rays backwards (dashed) - they meet on the object side, between focal point and centre. Always a virtual image.

Common mistake. Using 1/v + 1/u = 1/f (the mirror formula) instead of 1/v - 1/u = 1/f (the lens formula). The sign on 1/u flips between mirror and lens formulas in the NCERT convention. Memorise both separately.

Why the combination diverges even though one lens converges. Power is P = 1/f. The convex lens has P₁ = +1/0.30 = +3.33 D; the concave lens has P₂ = -1/0.20 = -5 D. Adding: P = 3.33 - 5 = -1.67 D ⇒ f = -0.60 m = -60 cm. The "stronger" lens (in terms of |P|) dominates the net behaviour.

Real-world link - achromatic doublets. Real telescope objectives combine a convex (crown) and a concave (flint) lens in contact to cancel chromatic aberration. The two lenses are chosen so their net focal length is finite (converging) while their net dispersion is zero. The combination formula is the starting point for that design.

Alternative - direct power addition. P₁ + P₂ = 3.33 - 5 = -1.67 D; f = 1/P = -60 cm. Faster than the reciprocal arithmetic.

Common mistake. Forgetting the sign of f₂ (the concave lens). Once you write 1/f = 1/30 + 1/(-20), the sign takes care of itself. Skipping the sign and writing 1/30 + 1/20 gives f = +12 cm - wrong by both magnitude and sign.

Why near-point setting gives bigger magnification. The eyepiece's "near-point" magnification is 1 + D/f_e, versus the "infinity" magnification D/f_e. The extra +1 gives a fractional boost ∼ f_e/D - small but noticeable. Trade-off: near-point viewing tires the eye ciliary muscles strain to focus at 25 cm; infinity viewing is more relaxing. Microscopists set their instrument for relaxed viewing during long observations.

Tube length and "tube length convention." Many textbooks use L ≈ tube length ≈ v_o because |u_o|≪ v_o. Approximate magnification:

m_{near point} ≈ -

L

(1+Df_e), m_∞ ≈ -Lf_o·Df_e.

f_o

NCERT often quotes these approximate formulas. They're accurate to a few percent.

Real-world link - modern microscopes. Today's compound microscopes use objective lenses with very short focal lengths ∼ 2 mm at 100× and a "tube lens" rather than a simple eyepiece. The intermediate image forms at a precise location, allowing for parallel beam paths between objective and tube lens - much easier to insert filters, beam splitters, etc.

Common mistake. Treating the eyepiece distance as v_e = +25 cm for near-point viewing. The virtual image is on the same side as the object of the eyepiece - so v_e = -25 cm in the Cartesian convention.

Why an 8-mm objective is so powerful. Magnification of the objective is v_o/u_o. Putting the object just outside the focal point here 0.9 cm vs. 0.8 cm means v_o is huge - small denominator changes blow up the ratio. That's the magic of compound microscopes: tiny focal lengths and near-focus object placement.

Trade-off - depth of field. Very short focal lengths give huge magnification but vanishingly small depth of field. At 100× total magnification you can focus only on a ∼ 1 -thick "slice" of the specimen - anything above or below is blurred. That's why you continually adjust the fine focus.

Real-world link - oil immersion. To boost resolution further, an "oil immersion" objective uses immersion oil between specimen and lens matching glass's refractive index ≈ 1.5. This eliminates refraction at the cover slip, increases the effective numerical aperture, and lets the objective resolve features down to ∼ 200 nm.

Common mistake. Confusing units - f_o = 8.0 mm, not 8.0 cm. Always convert to consistent units before substituting.

Why telescopes use a long objective and short eyepiece. Magnification = f_o/f_e. Big objective focal length gives a large real image of distant objects (think of it as a magnifying glass of the sky); a short-focal-length eyepiece then magnifies that image strongly. This is exactly the opposite of a microscope, which uses a short objective to get a tiny object to look big.

Alternative - near-point setting. If the eyepiece is adjusted so the final image forms at D = 25 cm instead of infinity,

|m|_nearpoint =

f_o

(1 + f_eD).

f_e

For this telescope: 24 × (1 + 6/25) = 24 × 1.24 = 29.76. Slightly more magnification, but the eye must accommodate.

Real-world link - light gathering. A telescope's most important property in astronomy isn't magnification but aperture (the diameter of the objective). A larger objective collects more light, allowing dim stars to be seen. The 25-m E-ELT in Chile collects ∼ 100,000× more light than the human eye.

Common mistake. Forgetting that telescope magnification is the ratio of angular sizes, not linear sizes. The image and object are at very different distances, so linear magnification is meaningless - angular magnification (i.e., the apparent size at the eye) is what matters.

Where the formula d = fθ comes from. For an object at infinity subtending angle θ, all rays from a single point hit the lens parallel. The lens focuses parallel rays at its second focal point. Two points on the object separated by angle θ thus form images separated by f tanθ ≈ fθ for small θ. Compact and powerful.

Why "refracting" vs "reflecting" telescopes. A 15-m focal length objective lens would weigh hundreds of kg - impractical. Modern large telescopes use a parabolic mirror instead, with the same focal length but a fraction of the weight. The Hubble Space Telescope uses a 2.4-m mirror; the James Webb uses a 6.5-m segmented mirror.

Real-world link - full-disk lunar imaging. Most amateur telescopes have a focal length of ∼ 1 m; the moon's image is about 0.9 cm across at the focal plane - fits comfortably on a small CCD chip. The 15-m telescope here produces a 13.7 cm image - needs a much larger detector (or special "focal reducer" optics to compress it).

Common mistake. Using sinθ instead of θ for the angular size. For θ less than about 5°, the difference is below 0.4% - but trigonometric calculators give sinθ very different from θ if you accidentally calculate in degrees. Always work in radians for small-angle formulas.

Pedagogical value of this exercise. Most students learn the six "standard cases" for a concave mirror by rote (object at infinity, beyond C, at C, between F and C, at F, between P and F). This question forces you to prove the image properties from the formula - which means once you've done it, you never need to memorise the cases.

Algebraic vs graphical methods. Each property here can also be verified by drawing ray diagrams. But ray diagrams suffer from "what if my drawing is off?" The algebra is exact. In JEE/NEET MCQs, sketching is faster; in long-answer questions, the algebra is required.

Common mistake - sign of f. Throughout this question, f is negative for concave mirrors and positive for convex mirrors. Mixing the sign halfway through the derivation produces nonsense conclusions like "convex mirrors give real images for nearby objects" - a result every student knows is wrong.

Real-world link - security mirrors. The combination "virtual, erect, diminished, wider field-of-view" makes convex mirrors ideal for shop ceilings, parking-lot exits, and roadside corners. Property (b) - that the image is always virtual - guarantees no one will see an upside-down reflection of themselves, which would be confusing.

Derivation of Δ = t1-1/μ. Inside the slab, the pin's image is "lifted" by an amount equal to (real depth in glass - apparent depth in glass) = t - t/μ = t1 - 1/μ. When you exit the slab back into air, no further shift occurs because the rays are now travelling in air at the same angle they had above the slab (the slab merely shifts the ray laterally, doesn't change its direction). Hence the total shift equals the in-slab shift - independent of slab position.

Why slab location is irrelevant - deeper reason. A slab of glass acts on light only inside its volume. The angle of the emerging ray equals the angle of the incident ray (lateral displacement, no angular change for parallel-sided slab). So as far as the observer is concerned, all that matters is how much "extra optical path" was traversed - and that depends only on t and μ, not on position.

Real-world link - viewing fish through aquarium glass. The fish appears slightly closer than it actually is, by an amount t1-1/μ. For a typical aquarium pane t = 1 cm, μ = 1.5, Δ = 0.33 cm - barely noticeable but real. Underwater photographers must account for this when calibrating focus.

Common mistake. Plugging the distance from eye to pin 50 cm into the formula instead of the slab thickness. The eye-to-pin distance is irrelevant to the apparent-shift formula - only the slab matters.