How to Use This Problem Set

Your full workout for Ray Optics, grouped by theme: mirrors, refraction and TIR, spherical surfaces and lenses, power and combinations, prisms, and instruments.

Keep these handy (Cartesian signs throughout):

  • Mirror: f=R/2f = R/2; 1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f}; m=−v/um = -v/u
  • Snell: sin⁡isin⁡r=n21\frac{\sin i}{\sin r} = n_{21}; n=c/vn = c/v; apparent depth = real/n; sin⁡ic=1/n\sin i_c = 1/n
  • Surface: n2v−n1u=n2−n1R\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}; Lens maker: 1f=(n21−1)(1R1−1R2)\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
  • Lens: 1v−1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f}; m=v/um = v/u; P=1/fP = 1/f (D); contact: P=∑PiP = \sum P_i, m=∏mim = \prod m_i
  • Prism: r1+r2=Ar_1+r_2 = A; d=i+e−Ad = i+e-A; n=sin⁡[(A+Dm)/2]sin⁡(A/2)n = \frac{\sin[(A+D_m)/2]}{\sin(A/2)}; thin: Dm=(n−1)AD_m = (n-1)A
  • Instruments: magnifier 1+D/f1 + D/f or D/fD/f; microscope LfoDfe\frac{L}{f_o}\frac{D}{f_e}; telescope fofe\frac{f_o}{f_e}, tube fo+fef_o+f_e

Sign first, substitute second.

Solved Examples - Mirrors

Example 1. Concave mirror, R = 36 cm: focal length?

Solution: f=R/2=−18f = R/2 = -18 cm (concave: negative).

Example 2. Object 24 cm before a concave mirror (f = −16-16 cm): image?

Solution: 1v=−116+124=−148\frac{1}{v} = -\frac{1}{16} + \frac{1}{24} = -\frac{1}{48}: v=−48v = -48 cm; m=−v/u=−2m = -v/u = -2 — real, inverted, doubled.

Example 3. Object 12 cm before the same mirror:

Solution: 1v=−116+112=148\frac{1}{v} = -\frac{1}{16} + \frac{1}{12} = \frac{1}{48}: v=+48v = +48 cm; m=+4m = +4 — virtual, erect, 4x (inside F).

Example 4. Convex mirror f = +20 cm, object at 30 cm:

Solution: 1v=120+130=112\frac{1}{v} = \frac{1}{20} + \frac{1}{30} = \frac{1}{12}: v=+12v = +12 cm; m=+0.4m = +0.4 — virtual, erect, diminished, as always.

Example 5. Where before a concave mirror (f = −10-10 cm) is the image real and 4x?

Solution: m=−4m = -4: v=4uv = 4u. Equation: 14u+1u=−110\frac{1}{4u} + \frac{1}{u} = -\frac{1}{10}: 54u=−110\frac{5}{4u} = -\frac{1}{10}, u=−12.5u = -12.5 cm.

Example 6. NCERT 9.4 anchor: convex mirror R = 2 m, jogger at 9 m moving 5 m/s — image speed?

Solution: From v = uf/(u-f): over 1 s, v goes 9/10 to 4/5 m: shift 1/10 m — image speed about 110\frac{1}{10} m/s.

Example 7. A mirror forms a real, same-size image at 40 cm. Its type and f?

Solution: m = −1-1 means v = u = −40-40 cm: object at C. Concave, f=−20f = -20 cm.

Solved Examples - Refraction & TIR

Example 8. i = 45 degrees, glass n = 1.414: r?

Solution: sin⁡r=0.7071.414=0.5\sin r = \frac{0.707}{1.414} = 0.5: r=30∘r = 30^{\circ}.

Example 9. Speed of light in diamond (n = 2.42)?

Solution: v=c/n=3×1082.42=1.24×108v = c/n = \frac{3\times10^8}{2.42} = 1.24\times10^8 m/s.

Example 10. A tank holds 2.0 m of water (n = 4/3). Apparent depth?

Solution: 2.04/3=1.5\frac{2.0}{4/3} = 1.5 m.

Example 11. Critical angle of diamond?

Solution: sin⁡ic=1/2.42=0.413\sin i_c = 1/2.42 = 0.413: ic=24.4∘i_c = 24.4^{\circ} — the sparkle's secret.

Example 12. Can a 45-degree TIR prism be made of material with n = 1.3?

Solution: ic=sin⁡−1(1/1.3)=50.3∘>45∘i_c = \sin^{-1}(1/1.3) = 50.3^{\circ} > 45^{\circ}: no — light at 45 degrees refracts out instead of totally reflecting.

Example 13. A slab (n = 1.5, t = 9 cm) lies over a dot. Apparent rise?

Solution: rise =t(1−1/n)=9×13=3= t(1 - 1/n) = 9 \times \frac{1}{3} = 3 cm.

Example 14. n31n_{31} = 1.6, n21n_{21} = 1.2: find n32n_{32}.

Solution: n32=n31/n21=1.6/1.2=1.33n_{32} = n_{31}/n_{21} = 1.6/1.2 = 1.33.

Example 15. Light from glass (n = 1.5) into air at i = 60 degrees (> ici_c = 41.8): result?

Solution: Total internal reflection — the surface acts as a perfect mirror; the ray reflects at 60 degrees.

Solved Examples - Surfaces & Lenses

Example 16. NCERT 9.5 anchor: air-to-glass (n = 1.5), R = +20 cm, u = −100-100 cm: v?

Solution: 1.5v=0.025−0.01=0.015\frac{1.5}{v} = 0.025 - 0.01 = 0.015: v=+100v = +100 cm.

Example 17. Lens maker: n = 1.5, R1=+20R_1 = +20, R2=−20R_2 = -20 cm: f?

Solution: 1f=0.5(120+120)=120\frac{1}{f} = 0.5\left(\frac{1}{20}+\frac{1}{20}\right) = \frac{1}{20}: f=+20f = +20 cm (equiconvex with n = 1.5: f = R).

Example 18. Object 15 cm from a convex lens f = +10 cm:

Solution: 1v=110−115=130\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}: v=+30v = +30 cm; m=−2m = -2 — real, inverted, doubled.

Example 19. Object 5 cm from the same lens:

Solution: 1v=110−15=−110\frac{1}{v} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10}: v=−10v = -10 cm; m=+2m = +2 — the magnifying-glass case.

Example 20. Concave lens f = −15-15 cm, object 30 cm away:

Solution: 1v=−115−130=−110\frac{1}{v} = -\frac{1}{15} - \frac{1}{30} = -\frac{1}{10}: v=−10v = -10 cm; m=+13m = +\frac{1}{3} — virtual, erect, diminished.

Example 21. A lens (n = 1.5, f = +12 cm in air) enters water (n = 4/3). New f?

Solution: Ratio trick: fwfa=(na−1)(nw−1)\frac{f_w}{f_a} = \frac{(n_a - 1)}{(n_w - 1)} with na=1.5n_a = 1.5, nw=1.54/3=1.125n_w = \frac{1.5}{4/3} = 1.125: fw12=0.50.125=4\frac{f_w}{12} = \frac{0.5}{0.125} = 4: fw=+48f_w = +48 cm.

Example 22. Sun's image by a convex lens f = +40 cm forms at:

Solution: Object at infinity: at the focus, 40 cm — the burn-spot distance.

Example 23. For a convex lens of f = +10 cm, where must the object sit to form a real, same-size image — and where does that image form?

Solution: u=−20u = -20 cm, v=+20v = +20 cm, m = −1-1 — the 2F-2F' symmetric pair.

Solved Examples - Power & Combinations

Example 24. f = +25 cm: power?

Solution: P=10.25=+4P = \frac{1}{0.25} = +4 D.

Example 25. P = −2.5-2.5 D: lens type and f?

Solution: Concave, f=−40f = -40 cm.

Example 26. +4 D and −1-1 D in contact: combined f?

Solution: P = +3 D: f=+13f = +\frac{1}{3} m ≈33.3\approx 33.3 cm.

Example 27. f1 = +20 and f2 = +30 cm in contact:

Solution: 1f=120+130=112\frac{1}{f} = \frac{1}{20}+\frac{1}{30} = \frac{1}{12}: f=+12f = +12 cm; P = 8.3 D.

Example 28. What lens in contact converts a +5 D lens into a +2 D system?

Solution: P2=2−5=−3P_2 = 2 - 5 = -3 D: a concave lens of f=−33.3f = -33.3 cm.

Solved Examples - Prisms

Example 29. A = 60 degrees, n = 1.5: find DmD_m.

Solution: sin⁡(60+Dm2)=1.5sin⁡30∘=0.75\sin\left(\frac{60+D_m}{2}\right) = 1.5\sin30^{\circ} = 0.75: 60+Dm2=48.6∘\frac{60+D_m}{2} = 48.6^{\circ}: Dm=37.2∘D_m = 37.2^{\circ}.

Example 30. Thin prism A = 5 degrees, n = 1.6: deviation?

Solution: D=(n−1)A=0.6×5=3∘D = (n-1)A = 0.6\times5 = 3^{\circ}.

Example 31. At minimum deviation through a 60-degree prism, the ray inside makes what angle with each face's normal?

Solution: r=A/2=30∘r = A/2 = 30^{\circ} at both faces — parallel to the base.

Solved Examples - Instruments

Example 32. Magnifier f = 2.5 cm: both magnifications (D = 25 cm)?

Solution: Near point: 1+252.5=111 + \frac{25}{2.5} = 11; infinity: 252.5=10\frac{25}{2.5} = 10.

Example 33. Microscope: fo=1.25f_o = 1.25 cm, fe=5f_e = 5 cm, L = 20 cm, image at infinity:

Solution: m=201.25×255=16×5=80m = \frac{20}{1.25}\times\frac{25}{5} = 16\times5 = 80.

Example 34. Telescope: fo=150f_o = 150 cm, fe=5f_e = 5 cm — magnification and tube length?

Solution: m=30m = 30; tube =155= 155 cm. (And swapping the lenses would shrink the world 30-fold.)