Optics
Optics for JEE Main & Advanced
Ray Optics
Reflection by MirrorsTopic 1
Laws of Reflection: (i) Angle of incidence = angle of reflection. (ii) Incident ray, reflected ray, normal lie in same plane.
Sign Convention (Cartesian): Distances measured from pole, in direction of incident light positive, against negative. Heights above principal axis positive, below negative.
Mirror Equation: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$, where $f$ = focal length, $u$ = object distance, $v$ = image distance.
Magnification: $m = -v/u = h'/h$. Positive: virtual & erect; negative: real & inverted.
Relation: $f = R/2$ for spherical mirrors ($R$ = radius of curvature).
Concave mirror ($f < 0$ in convention): produces real, inverted images for object beyond $F$; virtual, erect, magnified for object between $F$ and $P$.
Convex mirror ($f > 0$): always virtual, erect, diminished image (used as rear-view mirror).
Image Formation by Concave Mirror:
| Object position | Image position | Nature |
|---|---|---|
| At infinity | At $F$ | Real, point |
| Beyond $C$ | Between $F$ and $C$ | Real, inverted, diminished |
| At $C$ | At $C$ | Real, inverted, same size |
| Between $C$ and $F$ | Beyond $C$ | Real, inverted, magnified |
| At $F$ | At infinity | Real, highly magnified |
| Between $F$ and $P$ | Behind mirror | Virtual, erect, magnified |
For convex mirror: image is always virtual, erect, diminished, between $P$ and $F$.
An object is placed $30$ cm in front of a concave mirror of $f = 20$ cm. Find image position and magnification.
Show solution
$u = -30$, $f = -20$. $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{20} + \frac{1}{30} = \frac{-3+2}{60} = -\frac{1}{60}$$ $v = -60$ cm. Image $60$ cm in front (real). $m = -v/u = -60/30 = -2$.
Final Answer: Image at $60$ cm in front; $m = -2$ (real, inverted, magnified $2\times$).
A $5$ cm object is placed $10$ cm from convex mirror of $f = 15$ cm. Find image distance and size.
Show solution
$u = -10$, $f = +15$. $$\frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2+3}{30} = \frac{1}{6} \implies v = 6 \text{ cm}$$ $m = -v/u = -6/(-10) = 0.6$. Image size = $0.6 \times 5 = 3$ cm.
Final Answer: $v = +6$ cm (virtual, behind mirror); size $3$ cm, erect.
Focal length of a spherical mirror of $R = 20$ cm:
A convex mirror always forms:
The image formed by a plane mirror is:
In a concave mirror, when object is at $C$:
Magnification of mirror given by:
Refraction, Lenses and PrismTopic 2
Laws of Refraction (Snell's): $n_1\sin\theta_1 = n_2\sin\theta_2$. Refractive index: $n = c/v$ where $v$ = speed of light in medium.
Total Internal Reflection (TIR): Occurs when light travels from denser to rarer medium at angle $> \theta_c$ (critical angle): $$\sin\theta_c = \frac{1}{n} = \frac{n_2}{n_1} \quad (n_1 > n_2)$$ Examples: mirage, optical fibre, brilliance of diamond ($n = 2.42$, $\theta_c \approx 24°$).
Refraction at Spherical Surface: $$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$
Lens Maker's Formula: $$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $R_1, R_2$ with sign convention.
Thin Lens Formula: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$. Magnification: $m = v/u$. Power: $P = 1/f(\text{in m})$, units diopter (D).
Lens Combinations (in contact): $P = P_1 + P_2$, $1/f = 1/f_1 + 1/f_2$.
Prism:
- Deviation: $\delta = (i_1 + i_2) - A$ where $A$ = prism angle
- Minimum deviation when $i_1 = i_2$ ($r_1 = r_2 = A/2$)
- $n = \dfrac{\sin[(A+\delta_m)/2]}{\sin(A/2)}$
- For small prism: $\delta = (n-1)A$
Dispersion: Splitting of white light into colors by prism. Dispersive power: $\omega = (\mu_v - \mu_r)/(\mu - 1)$.
Achromatic combination of prisms: $\omega_1 A_1 + \omega_2 A_2 = 0$ (chosen so dispersions cancel but deviation remains).
Light enters from water ($n = 1.33$) into glass ($n = 1.5$) at $30°$. Find angle of refraction.
Show solution
$$n_1\sin\theta_1 = n_2\sin\theta_2$$ $$\sin\theta_2 = \frac{1.33 \times 0.5}{1.5} = 0.4433$$ $\theta_2 = \sin^{-1}(0.4433) \approx 26.3°$.
Final Answer: $\theta_2 \approx 26.3°$.
A biconvex lens has $R_1 = 20$ cm, $R_2 = -30$ cm, $n = 1.5$. Find focal length.
Show solution
$$\frac{1}{f} = (1.5-1)\left(\frac{1}{20} - \frac{1}{-30}\right) = 0.5\left(\frac{1}{20} + \frac{1}{30}\right)$$ $$= 0.5 \times \frac{3+2}{60} = 0.5 \times \frac{5}{60} = \frac{1}{24}$$ $f = 24$ cm.
Final Answer: $f = 24$ cm.
Critical angle for glass ($n = 1.5$):
The power of a lens of focal length $-20$ cm:
Total internal reflection occurs when light goes:
Two thin lenses of $f_1 = 10$ cm and $f_2 = -15$ cm in contact. Equivalent focal length:
White light passing through a prism splits into colours because of:
Wave Optics
Interference and YDSETopic 1
Huygens' Principle: Every point on a wavefront is a source of secondary wavelets; the new wavefront is the envelope of these.
Coherent Sources: Sources of same frequency, fixed phase relation, comparable amplitude. Required for sustained interference. Typically derived from same source (Young's slits, Lloyd's mirror, Fresnel biprism).
Superposition / Interference: Resultant intensity at point with phase difference $\phi$: $$I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$$
For $I_1 = I_2 = I_0$: $I = 4I_0\cos^2(\phi/2)$.
- Constructive interference: $\phi = 2n\pi$ (path difference $= n\lambda$); $I_{\max} = (\sqrt{I_1}+\sqrt{I_2})^2$
- Destructive interference: $\phi = (2n+1)\pi$ (path difference $= (n+1/2)\lambda$); $I_{\min} = (\sqrt{I_1}-\sqrt{I_2})^2$
Young's Double Slit Experiment (YDSE): Two slits separated by $d$, screen at distance $D$.
| Quantity | Formula |
|---|---|
| Path difference at point $y$ | $\Delta x = yd/D$ |
| Bright fringe positions | $y_n = n\lambda D/d$, $n = 0, \pm 1, \pm 2,\ldots$ |
| Dark fringe positions | $y_n = (n+1/2)\lambda D/d$ |
| Fringe width | $\beta = \lambda D/d$ |
| Angular fringe width | $\theta = \lambda/d$ |
Fringe width independent of fringe order. Decreases if $d$ increases or $D$ decreases.
In medium of refractive index $n$: $\lambda' = \lambda/n$, so fringe width decreases by factor $n$.
Effect of inserting glass slab of thickness $t$ and index $n$ in one path: extra path = $(n-1)t$ → pattern shifts by $(n-1)tD/d$.
In YDSE, $\lambda = 500$ nm, $d = 1$ mm, $D = 1$ m. Find fringe width.
Show solution
$$\beta = \frac{\lambda D}{d} = \frac{500 \times 10^{-9} \times 1}{10^{-3}} = 5 \times 10^{-4} \text{ m} = 0.5 \text{ mm}$$
Final Answer: $\beta = 0.5$ mm.
In YDSE the entire arrangement is immersed in water ($n = 4/3$). If original $\beta = 0.6$ mm, find new $\beta$.
Show solution
$$\beta' = \beta/n = 0.6/(4/3) = 0.45 \text{ mm}$$
Final Answer: $\beta' = 0.45$ mm.
For sustained interference, sources must be:
Fringe width in YDSE is proportional to:
At a point of destructive interference, intensity:
If YDSE is performed in vacuum versus water, fringe width:
The central fringe in YDSE is:
Diffraction, Polarization and Optical InstrumentsTopic 2
Diffraction: Bending of light around obstacles or through narrow openings. Significant when slit width $\sim \lambda$.
Single Slit Diffraction:
- Position of dark fringes: $d\sin\theta = n\lambda$, $n = \pm 1, \pm 2, \ldots$
- Position of secondary maxima: $d\sin\theta \approx (n + 1/2)\lambda$
- Width of central maximum: $2\lambda D/d$ (twice the width of secondary maxima)
Resolving Power:
- Microscope: $R = 2n\sin\theta/(1.22\lambda)$
- Telescope: $R = D/(1.22\lambda)$, where $D$ = aperture
Larger aperture / shorter wavelength → better resolution.
Polarization: EM waves are transverse → can be polarized. Unpolarized light has $\vec{E}$ vibrating in all directions; polarized has it in one plane.
Malus' Law: $I = I_0\cos^2\theta$, where $\theta$ = angle between polarizer axis and $\vec{E}$.
Brewster's Law: When light strikes a dielectric at $\theta_B$ such that reflected and refracted rays are perpendicular, the reflected ray is fully polarized. $$\tan\theta_B = n$$ (Brewster's angle).
Optical Instruments:
| Instrument | Magnification (image at $D$) | Magnification (image at infinity) |
|---|---|---|
| Simple microscope | $m = 1 + D/f$ | $m = D/f$ |
| Compound microscope | $m = (L/f_o)(1 + D/f_e)$ | $m = (L/f_o)(D/f_e)$ |
| Telescope | $m = (f_o/f_e)(1 + f_e/D)$ | $m = f_o/f_e$ |
| Length of telescope | — | $L = f_o + f_e$ |
For telescope: large $f_o$ and small $f_e$ for high magnification.
Unpolarized light of intensity $I_0$ passes through two polarizers at $60°$. Find transmitted intensity.
Show solution
After first polarizer: $I_1 = I_0/2$. After second (Malus' law): $I = I_1\cos^2 60° = (I_0/2)(1/4) = I_0/8$.
Final Answer: $I = I_0/8$.
A telescope has $f_o = 100$ cm, $f_e = 5$ cm. Find magnification and length (image at infinity).
Show solution
$$m = f_o/f_e = 100/5 = 20$$ $$L = f_o + f_e = 105 \text{ cm}$$
Final Answer: $m = 20$; $L = 105$ cm.
The width of central maximum in single-slit diffraction is:
Brewster's angle for water ($n = 1.33$):
Malus' law:
Resolving power of telescope depends on:
In compound microscope, the eyepiece has:
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