Optics — Easy Practice (Set 1)

Quick Recap — Reflection & Refraction

• Spherical mirror: $f = \dfrac{R}{2}$; mirror formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$; magnification $m = -\dfrac{v}{u}$.
• Refraction (Snell's law): $n_1\sin\theta_1 = n_2\sin\theta_2$; refractive index $n = \dfrac{c}{v}$.
• Total internal reflection (denser to rarer medium): critical angle $\sin C = \dfrac{1}{n}$.
• Apparent depth $= \dfrac{\text{real depth}}{n}$.
• Concave mirrors converge (shaving mirrors, telescopes); convex mirrors diverge (rear-view mirrors), always giving virtual, diminished images.

Beyond-NCERT JEE Formulae

A consolidated formula sheet for the calculation-heavy optics that sits beyond the basic mirror and lens equations -- silvered lenses, prisms, total internal reflection, the Young's double-slit toolkit, and resolving and magnifying power -- with when-to-use notes and JEE traps. The practice sets stated the one-line versions; here we gather them, add the beyond-NCERT forms, and flag where marks slip away.

1. Lens maker, power and combinations

• Lens maker's formula: $\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, where $n$ is the lens index relative to its surroundings. Immersed in a medium of index $n_m$, replace $(n-1)$ by $\left(\dfrac{n_{lens}}{n_m}-1\right)$, so $f$ lengthens and can even change sign.
• Power: $P=\dfrac{1}{f}$ in dioptre when $f$ is in metre; a positive $P$ converges, a negative $P$ diverges.
• Thin lenses in contact: $P=P_1+P_2+\cdots$, that is $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$.
• Lenses separated by $d$: $P=P_1+P_2-d\,P_1P_2$ with $d$ in metre; separation always weakens a converging pair.
• Silvered lens as an equivalent mirror: $\dfrac{1}{F}=\dfrac{2}{f_l}+\dfrac{1}{f_m}$, i.e. power $P=2P_l+P_m$, with $f_m=\dfrac{R}{2}$ for the silvered surface treated as a mirror; then finish with $\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{F}$.
• When to use: lens maker's for finding $f$ from $R$ and $n$ and for every immersion problem; the separated-lens power for coaxial two-lens systems; the silvered-lens rule the moment any face is mirror-coated.
• [JEE Tip] Light crosses a silvered lens twice, so the lens power enters twice as $2P_l$. The classic slip is taking the reflecting surface as $f_m=R$ instead of $\dfrac{R}{2}$.

2. Prism

• Thin-prism deviation: $\delta=(n-1)A$ for a small refracting angle $A$.
• General ray: $r_1+r_2=A$ and $\delta=i+e-A$.
• Minimum deviation (symmetric ray, $i=e$): $n=\dfrac{\sin\dfrac{A+\delta_m}{2}}{\sin\dfrac{A}{2}}$, with incidence $i=\dfrac{A+\delta_m}{2}$.
• Angular dispersion: $\theta=\delta_v-\delta_r=(n_v-n_r)A$.
• Dispersive power: $\omega=\dfrac{n_v-n_r}{n-1}$, with mean index $n=\dfrac{n_v+n_r}{2}$.
• Combined thin prisms: deviation without dispersion (direct vision) needs $(n_v-n_r)A+(n_v'-n_r')A'=0$; dispersion without deviation needs $(n-1)A+(n'-1)A'=0$.
• When to use: $\delta=(n-1)A$ for spectrometer and prism-combination sums; the min-deviation formula whenever a symmetric passage or $\delta_m$ is quoted.
• [JEE Tip] Immerse a prism and the formula switches to the relative index $\dfrac{n_{glass}}{n_m}$; the deviation then collapses, because the bending depends on how much denser the glass is than its surroundings.

3. Total internal reflection and optical fibre

• Critical angle: $\sin C=\dfrac{1}{n}$ for a dense medium of index $n$ against air; in general $\sin C=\dfrac{n_{rarer}}{n_{denser}}$.
• Escape cone: a source at depth $h$ lights a surface circle of radius $r=h\tan C$; outside it every ray is totally reflected.
• Optical-fibre numerical aperture: $NA=\sqrt{n_1^{2}-n_2^{2}}$ for core $n_1$ and cladding $n_2$; the acceptance half-angle in air obeys $\sin\theta_{max}=NA$.
• When to use: $\sin C=\dfrac{1}{n}$ for whether a ray escapes; the escape-cone radius for illuminated-patch problems; the numerical aperture for fibre acceptance cones.
• [JEE Tip] The numerical aperture is $\sqrt{n_1^{2}-n_2^{2}}$, never $n_1-n_2$; the raw difference underestimates the cone many times over.

4. Wave optics: Young's double slit

• Fringe width: $\beta=\dfrac{\lambda D}{d}$, the same spacing for bright and dark fringes.
• Path difference at height $y$: $\Delta=\dfrac{yd}{D}$, with phase $\phi=\dfrac{2\pi}{\lambda}\Delta$.
• Bright and dark: maxima at $\Delta=m\lambda$, minima at $\Delta=\left(m+\dfrac{1}{2}\right)\lambda$, for $m=0,1,2,\ldots$
• Intensity: $I=I_{max}\cos^{2}\dfrac{\phi}{2}$; for unequal sources $I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$.
• Thin slab over one slit: the pattern shifts by $\Delta y=\dfrac{(n-1)tD}{d}$, that is $\dfrac{(n-1)t}{\lambda}$ fringes toward that slit.
• Whole set-up immersed: $\lambda_m=\dfrac{\lambda}{n}$, so every fringe narrows by the factor $n$.
• When to use: $\beta=\dfrac{\lambda D}{d}$ for spacing; the slab formula when a sheet covers one slit; $\lambda_m=\dfrac{\lambda}{n}$ once the apparatus is under water.
• [JEE Tip] The slab shift $\dfrac{(n-1)tD}{d}$ carries no $\lambda$, so in white light the central band stays white at its new position; only the coloured side fringes wash out.

5. Resolving power and magnifying power

• Telescope (Rayleigh): limit of resolution $\Delta\theta=\dfrac{1.22\lambda}{D}$ set by the aperture $D$; resolving power $=\dfrac{D}{1.22\lambda}$.
• Microscope: smallest resolved separation $d_{min}=\dfrac{0.61\lambda}{NA}$ with $NA=n\sin\theta$; oil immersion raises $NA$ and sharpens the detail.
• Compound microscope: $M=m_o\times m_e$; final image at the near point gives $M=\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$, at infinity $M=\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$, with tube length $L$ and $D=25$ cm.
• Astronomical telescope: normal adjustment $M=\dfrac{f_o}{f_e}$ and length $L=f_o+f_e$; final image at the near point $M=\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$.
• Simple microscope: $M=1+\dfrac{D}{f}$ (image at the near point) or $\dfrac{D}{f}$ (image at infinity).
• When to use: the Rayleigh $\dfrac{1.22\lambda}{D}$ for just-resolved star or headlamp questions; $\dfrac{f_o}{f_e}$ for telescope power; the $1+\dfrac{D}{f_e}$ eyepiece factor only when the final image sits at the near point.
• [JEE Tip] Resolving power lives on the aperture $D$; magnifying power lives on the focal lengths. Swapping the two is the single most common telescope mistake.

Solved Examples -- Beyond-NCERT Formulae

Example 1 -- Lens maker's formula and immersion. A thin biconvex lens of glass index $1.5$ has radii $R_1=+15$ cm and $R_2=-30$ cm. Find its focal length and power in air, and its focal length in water of index $\dfrac{4}{3}$.

• In air: $\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=(0.5)\left(\dfrac{1}{15}+\dfrac{1}{30}\right)=(0.5)(0.1)=0.05$, so $f=20$ cm, i.e. $0.20$ m, and $P=\dfrac{1}{0.20}=+5$ D.
• In water: the factor becomes $\dfrac{1.5}{4/3}-1=0.125$, so $\dfrac{1}{f_w}=(0.125)(0.1)=0.0125$ and $f_w=80$ cm.

Immersion quarters the power, from $+5$ D to $+1.25$ D, because the glass is only slightly denser than water.

Example 2 -- Power of lens combinations. Two thin convex lenses have focal lengths $20$ cm and $40$ cm, so $P_1=5$ D and $P_2=2.5$ D. Find the power and equivalent focal length when they are (a) in contact and (b) coaxial and $20$ cm apart.

• In contact: $P=P_1+P_2=7.5$ D, so $F=\dfrac{1}{7.5}=0.133$ m, i.e. $13.3$ cm.
• Separated by $d=0.20$ m: $P=P_1+P_2-d\,P_1P_2=7.5-(0.20)(5)(2.5)=5$ D, so $F=\dfrac{1}{5}=0.20$ m, i.e. $20$ cm.

The gap drops the power from $7.5$ D to $5$ D: separating two converging lenses always weakens them.

Example 3 -- Silvered lens as an equivalent mirror. An equiconvex lens of index $1.5$ with each radius $20$ cm is silvered on one curved face. An object sits on the axis $10$ cm in front of the clear face. Locate the image.

• Lens power: $\dfrac{1}{f_l}=(n-1)\dfrac{2}{R}=(0.5)\dfrac{2}{20}=0.05$, so $f_l=20$ cm.
• Silvered face as a mirror: $f_m=\dfrac{R}{2}=10$ cm.
• Equivalent mirror: $\dfrac{1}{F}=\dfrac{2}{f_l}+\dfrac{1}{f_m}=\dfrac{2}{20}+\dfrac{1}{10}=0.20$, so $F=5$ cm (concave).
• Imaging with $f=-5$ cm and $u=-10$ cm: $\dfrac{1}{v}=\dfrac{1}{-5}-\dfrac{1}{-10}=-0.10$, so $v=-10$ cm and $m=-\dfrac{v}{u}=-1$.

The object at $2F$ gives a real, inverted, same-size image at $2F$: the silvered lens is just a concave mirror of focal length $5$ cm.

Example 4 -- Prism at minimum deviation. A glass prism of refracting angle $A=60^\circ$ is made of glass of index $1.6$. Find the angle of minimum deviation and the incidence angle at which it occurs.

• Minimum-deviation relation: $\sin\dfrac{A+\delta_m}{2}=n\sin\dfrac{A}{2}=1.6\sin30^\circ=0.8$.
• So $\dfrac{A+\delta_m}{2}=\sin^{-1}(0.8)=53.1^\circ$, giving $\delta_m=2(53.1^\circ)-60^\circ=46.3^\circ$.
• At minimum deviation the passage is symmetric, so the incidence angle is $i=\dfrac{A+\delta_m}{2}=53.1^\circ$.

Example 5 -- Dispersive power. A thin prism of refracting angle $A=6^\circ$ is made of flint glass with $n_v=1.66$ and $n_r=1.62$. Find the mean deviation, the angular dispersion, and the dispersive power.

• Mean index: $n=\dfrac{n_v+n_r}{2}=1.64$, so the mean deviation is $\delta=(n-1)A=(0.64)(6^\circ)=3.84^\circ$.
• Angular dispersion: $\theta=(n_v-n_r)A=(0.04)(6^\circ)=0.24^\circ$.
• Dispersive power: $\omega=\dfrac{n_v-n_r}{n-1}=\dfrac{0.04}{0.64}=0.0625$, which also equals $\dfrac{\theta}{\delta}$.

Example 6 -- Critical angle and escape cone. A small lamp lies $3$ m deep in a liquid of index $1.25$. Find the critical angle at the liquid-air surface and the radius of the bright circle of escaping light seen from above.

• Critical angle: $\sin C=\dfrac{1}{n}=\dfrac{1}{1.25}=0.8$, so $C=53.1^\circ$.
• Escape cone: light emerges only within a cone of half-angle $C$, so the lit circle has radius $r=h\tan C=3\times\dfrac{0.8}{0.6}=3\times1.333=4.0$ m.

Outside this $4.0$ m circle the underside of the surface looks silvery, the everyday signature of total internal reflection.

Example 7 -- Fringe width and slab shift. In a Young's set-up the slits are $1$ mm apart, the screen is $2$ m away, and $\lambda=600$ nm. Find the fringe width, then the shift of the central fringe when a sheet of index $1.5$ and thickness $6$ micrometre covers the upper slit.

• Fringe width: $\beta=\dfrac{\lambda D}{d}=\dfrac{(600\times10^{-9})(2)}{1\times10^{-3}}=1.2\times10^{-3}$ m, i.e. $1.2$ mm.
• Slab shift: the covered arm gains extra optical path $(n-1)t$, moving the centre by $\dfrac{(n-1)t}{\lambda}=\dfrac{(0.5)(6\times10^{-6})}{600\times10^{-9}}=5$ fringes.
• Hence $\Delta y=5\beta=6.0$ mm toward the upper (covered) slit.

Example 8 -- Telescope magnification and resolving power. An astronomical telescope in normal adjustment has an objective of focal length $100$ cm and aperture $10$ cm and an eyepiece of focal length $2$ cm, used with light of wavelength $500$ nm. Find its magnifying power, tube length, and smallest resolvable angular separation.

• Magnifying power: $M=\dfrac{f_o}{f_e}=\dfrac{100}{2}=50$.
• Tube length: $L=f_o+f_e=100+2=102$ cm.
• Rayleigh limit (uses the aperture, not the focal length): $\Delta\theta=\dfrac{1.22\lambda}{D}=\dfrac{1.22(500\times10^{-9})}{0.10}=6.1\times10^{-6}$ rad.