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📘 Ray Optics — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Ray Optics for JEE Main with previous year questions on reflection, refraction, lenses, and prisms. Each solution is explained in a simple, step-by-step manner to boost your score.
Ray Optics — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (4 April Shift 1)
When an object is placed 40 cm away from a spherical mirror an image of magnification \(1/2\) is produced. To obtain an image with magnification of \(1/3\), the object is to be moved :
(1) 40 cm away from the mirror (2) 80 cm away from the mirror
(3) 20 cm towards the mirror (4) 20 cm away from the mirror
(1) 40 cm away from the mirror (2) 80 cm away from the mirror
(3) 20 cm towards the mirror (4) 20 cm away from the mirror
✅ Answer & Explanation:
Correct Option: (1)
- Magnification for mirror \(m = \frac{f}{f-u}\).
- Case 1: \(\frac{1}{2} = \frac{f}{f - (-40)} \implies f+40 = 2f \implies f = 40 \, \text{cm}\) (Convex mirror).
- Case 2: \(\frac{1}{3} = \frac{40}{40 - u'} \implies 40 - u' = 120 \implies u' = -80 \, \text{cm}\).
- The object must be moved from 40 cm to 80 cm, which is 40 cm away.
Q2. JEE Main 2025 (23 Jan Shift 2)
A concave mirror of focal length f in air is dipped in a liquid of refractive index \(\mu\). Its focal length in the liquid will be :
(1) \(\mu f\) (2) \(f\)
(3) \(\frac{f}{\mu-1}\) (4) \(\frac{f}{\mu}\)
(1) \(\mu f\) (2) \(f\)
(3) \(\frac{f}{\mu-1}\) (4) \(\frac{f}{\mu}\)
✅ Answer & Explanation:
Correct Option: (2)
The focal length of a mirror depends only on its radius of curvature (\(f = \frac{R}{2}\)) and not on the surrounding medium. Therefore, it remains \(f\).
The focal length of a mirror depends only on its radius of curvature (\(f = \frac{R}{2}\)) and not on the surrounding medium. Therefore, it remains \(f\).
Q3. JEE Main 2025 (22 Jan Shift 1)
A car uses a convex mirror (\(R = 2 \, \text{m}\)). Another car approaches from behind at \(90 \, \text{km/hr} = 25 \, \text{m/s}\). When it is at distance of 24 m, the magnitude of acceleration of image is \(a\). The value of \(100a\) is :
✅ Answer & Explanation:
Correct Answer: 8
- Image velocity \(v_i = m^2 v_o = (\frac{f}{f-u})^2 v_o\).
- Acceleration \(a_i = \frac{dv_i}{dt} = \frac{d}{dt} [ (\frac{f}{f-u})^2 v_o ] = 2 v_o \frac{f^2}{(f-u)^3} \times (- \frac{du}{dt})\)?
- Actually, \(a_i = 2 v_o^2 \frac{f^2}{(f-u)^3}\).
- Substituting \(f=1, u=-24, v_o=25\):
- \(a_i = 2 \times 625 \times \frac{1}{(25)^3} = \frac{1250}{15625} = 0.08 \, \text{m/s}^2\).
- \(100a = \mathbf{8}\).
Q4. JEE Main 2025 (2 April Shift 2)
Objects in front of convex and concave mirrors, \(R=12, u=18\). Ratio of magnification \(m_{convex}/m_{concave}\) is :
(1) 1/2 (2) 2 (3) 3 (4) 1/3
(1) 1/2 (2) 2 (3) 3 (4) 1/3
✅ Answer & Explanation:
Correct Option: (1)
- \(f = \frac{R}{2} = 6 \, \text{cm}\). \(u = -18 \, \text{cm}\).
- Convex: \(m_1 = \frac{6}{6 - (-18)} = \frac{6}{24} = \frac{1}{4}\).
- Concave: \(m_2 = \frac{-6}{-6 - (-18)} = -\frac{6}{12} = -\frac{1}{2}\).
- Ratio \(|m_1|/|m_2| = (\frac{1}{4}) / (\frac{1}{2}) = \mathbf{\frac{1}{2}}\).
Q5. JEE Main 2021 (31 Aug Shift 1)
Mirrors \(M_1 \perp M_2\). Source P at (a, 2a). Shortest dist between images?
✅ Answer & Explanation:
Correct Option: (3)
- Images are at \(I_1(-a, 2a)\) and \(I_2(a, -2a)\).
- Dist \(= \sqrt{(2a)^2 + (4a)^2} = a\sqrt{20} = 2a\sqrt{5}\).
- Given \(\sqrt{5} = 2.3\), dist \(= 2a(2.3) = 4.6a\).
Q6. JEE Main 2025 (8 April Shift 2)
A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm, respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be :
(1) \(\frac{500}{11} \, \text{cm}\) (2) \(\frac{800}{11} \, \text{cm}\)
(3) \(\frac{700}{11} \, \text{cm}\) (4) \(\frac{600}{11} \, \text{cm}\)
(1) \(\frac{500}{11} \, \text{cm}\) (2) \(\frac{800}{11} \, \text{cm}\)
(3) \(\frac{700}{11} \, \text{cm}\) (4) \(\frac{600}{11} \, \text{cm}\)
✅ Answer & Explanation:
Correct Option: (4)
- The system consists of two lenses in contact:
- Liquid lens (Plano-concave): \(R_1 = \infty, R_2 = 30 \, \text{cm}\).
- \(1/f_l = (\mu_l - 1)(1/\infty - \frac{1}{30}) = (1.3 - 1)(-\frac{1}{30}) = -\frac{0.3}{30} = -\frac{1}{100}\).
- Glass lens (Concave-convex): \(R_1 = 30 \, \text{cm}, R_2 = 20 \, \text{cm}\).
- \(1/f_g = (\mu_g - 1)(\frac{1}{30} - \frac{1}{20}) = (1.5 - 1)(-\frac{10}{600}) = -\frac{0.5}{60} = -\frac{1}{120}\).
- Combination: \(\frac{1}{f} = 1/f_l + 1/f_g = -\frac{1}{100} - \frac{1}{120} = \frac{-6-5}{600} = -\frac{11}{600}\).
- Magnitude of focal length = 600/11 cm.
Q7. JEE Main 2025 (7 April Shift 1)
Two thin convex lenses of focal length 30 cm and 10 cm are placed coaxially, 10 cm apart. The power of this combination is :
(1) 5 D (2) 1 D (3) 20 D (4) 10 D
(1) 5 D (2) 1 D (3) 20 D (4) 10 D
✅ Answer & Explanation:
Correct Option: (4)
- Equivalent focal length: \(\frac{1}{f} = 1/f_1 + 1/f_2 - d/(f_1 f_2)\).
- \(\frac{1}{f} = \frac{1}{30} + \frac{1}{10} - 10/(30 \times 10) = \frac{1}{30} + \frac{3}{30} - \frac{1}{30} = \frac{3}{30} = \frac{1}{10}\).
- \(f = 10 \, \text{cm} = 0.1 \, \text{m}\).
- Power \(P = \frac{1}{f}(m) = \frac{1}{0.1} = \mathbf{10 \, D}\).
Q8. JEE Main 2025 (29 Jan Shift 2)
A convex lens made of glass (\(\mu = 1.5\)) has focal length 24 cm in air. When it is totally immersed in water (\(\mu = 1.33 \approx 4/3\)), its focal length changes to :
(1) 24 cm (2) 96 cm (3) 48 cm (4) 72 cm
(1) 24 cm (2) 96 cm (3) 48 cm (4) 72 cm
✅ Answer & Explanation:
Correct Option: (2)
- \(f_{liq} = f_{air} \times \frac{(\mu_g - 1)}{(\mu_g/\mu_l - 1)}\).
- \(f_w = 24 \times \frac{(1.5 - 1)}{(\frac{1.5}{1.33} - 1)} = 24 \times \frac{0.5}{0.125} = 24 \times 4 = \mathbf{96 \, \text{cm}}\).
Q9. JEE Main 2025 (29 Jan Shift 1)
Let u and v be the distances of the object and the image from a lens of focal length f. The correct graphical representation of u and v for a convex lens when \(|u| > f\), is :
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (3)
For a convex lens, as the object moves from infinity towards the focus (\(|u| > f\)), the real image moves from focus towards infinity. The relationship \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\) produces a hyperbola in the u-v plane. Graph (3) correctly illustrates this asymptotic behavior.
For a convex lens, as the object moves from infinity towards the focus (\(|u| > f\)), the real image moves from focus towards infinity. The relationship \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\) produces a hyperbola in the u-v plane. Graph (3) correctly illustrates this asymptotic behavior.
Q10. JEE Main 2025 (28 Jan Shift 2)
In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '-2', then the radius of curvature of meniscus is :
(1) \(\frac{1}{3} \, \text{cm}\) (2) \(\frac{4}{3} \, \text{cm}\)
(3) \(1 \, \text{cm}\) (4) \(\frac{2}{3} \, \text{cm}\)
(1) \(\frac{1}{3} \, \text{cm}\) (2) \(\frac{4}{3} \, \text{cm}\)
(3) \(1 \, \text{cm}\) (4) \(\frac{2}{3} \, \text{cm}\)
✅ Answer & Explanation:
Correct Option: (4)
- For spherical refraction: \(m = \frac{\mu_1 v}{\mu_2 u}\).
- \(-2 = \frac{1.3 v}{1.4 (-13)} \implies v = \frac{2 \times 1.4 \times 13}{1.3} = 28 \, \text{cm}\).
- Now use \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\):
- \(\frac{1.4}{28} - \frac{1.3}{-13} = \frac{1.4 - 1.3}{R} \implies 0.05 + 0.1 = \frac{0.1}{R}\).
- \(0.15 = \frac{0.1}{R} \implies R = 0.1 / 0.15 = \frac{2}{3} \, \text{cm}\).
Q11. JEE Main 2025 (24 Jan Shift 1)
A thin plano-convex lens (\(\mu_g = 1.5\)) is immersed in a liquid (\(\mu_l = 1.2\)). When the plane side is silver coated, it behaves like a concave mirror of focal length 0.2 m. The radius of curvature of the curved surface is :
(1) 0.20 m (2) 0.25 m (3) 0.15 m (4) 0.10 m
(1) 0.20 m (2) 0.25 m (3) 0.15 m (4) 0.10 m
✅ Answer & Explanation:
Correct Option: (4)
- Equivalent focal length \(F\) of silvered lens: \(\frac{1}{F} = 2/f_l + 1/f_m\).
- Plane mirror \(f_m = \infty\), so \(\frac{1}{F} = 2/f_l\).
- Focal length in liquid: \(1/f_l = (\frac{\mu_g}{\mu_l} - 1)(\frac{1}{R} - \frac{1}{\infty}) = (\frac{1.5}{1.2} - 1)\frac{1}{R} = \frac{0.25}{R}\).
- \(\frac{1}{0.2} = 2 \times (\frac{0.25}{R}) = \frac{0.5}{R} \implies R = 0.5 \times 0.2 = \mathbf{0.10 \, \text{m}}\).
Q12. JEE Main 2025 (23 Jan Shift 1)
What is the lateral shift of a ray refracted through a parallel-sided glass slab of thickness 'h' in terms of angle of incidence 'i' and angle of refraction 'r'?
(1) \(h \frac{\tan(i-r)}{\tan r}\) (2) \(h \frac{\sin(i-r)}{\cos r}\)
(3) \(h\) (4) \(h \frac{\cos(i-r)}{\sin r}\)
(1) \(h \frac{\tan(i-r)}{\tan r}\) (2) \(h \frac{\sin(i-r)}{\cos r}\)
(3) \(h\) (4) \(h \frac{\cos(i-r)}{\sin r}\)
✅ Answer & Explanation:
Correct Option: (2)
- Standard derivation for lateral shift \(x\) through a slab of thickness \(t\):
- \(x = t \frac{\sin(i-r)}{\cos r}\). In this problem, thickness is denoted as \(h\).
Q13. JEE Main 2025 (22 Jan Shift 2)
A transparent film (\(\mu = 2.0\)) is coated on glass (\(\mu = 1.45\)). Min thickness for maximum transmission of Green light (\(\lambda = 550 \, \text{nm}\)) is :
(1) 137.5 nm (2) 275 nm (3) 94.8 nm (4) 68.7 nm
(1) 137.5 nm (2) 275 nm (3) 94.8 nm (4) 68.7 nm
✅ Answer & Explanation:
Correct Option: (1)
- Maximum transmission occurs when reflections from top and bottom surfaces interfere destructively.
- Since \(1 < 2.0 > 1.45\), only the first reflection has a \(\pi\) phase shift.
- Destructive interference condition: \(2\mu t = \lambda/2\).
- \(t = \lambda / (4\mu) = 550 / (4 \times 2) = 550 / 8 = \mathbf{137.5 \, \text{nm}}\).
Q14. JEE Main 2025 (22 Jan Shift 1)
In the diagram, three lenses are formed (water-glass-water). The power of the combination is :
(1) \(\frac{1}{6} (\frac{1}{|R_1|} - \frac{1}{|R_2|})\) (2) \(-\frac{1}{6} (\frac{1}{|R_1|} + \frac{1}{|R_2|})\)
(3) \(\frac{1}{6} (\frac{1}{|R_1|} + \frac{1}{|R_2|})\) (4) \(-\frac{1}{6} (\frac{1}{|R_1|} - \frac{1}{|R_2|})\)
(3) \(\frac{1}{6} (\frac{1}{|R_1|} + \frac{1}{|R_2|})\) (4) \(-\frac{1}{6} (\frac{1}{|R_1|} - \frac{1}{|R_2|})\)
✅ Answer & Explanation:
Correct Option: (4)
- Total power \(P = \sum (\mu_2 - \mu_1)/R\).
- Summing the refractive interfaces for the stacked water, glass, and water layers leads to the result in Option (4).
Q15. JEE Main 2025 (22 Jan Shift 1)
Thin convex lens glass (\(\mu, R\)). One side polished. At what distance should object be placed so image forms on self?
(1) \(R/\mu\) (2) \(R/(2\mu-3)\)
(3) \(\mu R\) (4) \(R/(2\mu-1)\)
(1) \(R/\mu\) (2) \(R/(2\mu-3)\)
(3) \(\mu R\) (4) \(R/(2\mu-1)\)
✅ Answer & Explanation:
Correct Option: (4)
- Image forms on object if rays hit the silvered surface normally.
- Equivalent focal length \(F = R / [2(2\mu-1)]\).
- Object distance \(u = 2F = \mathbf{R / (2\mu - 1)}\).
Q16. JEE Main 2025 (2 April Shift 1)
Spherical surface, indices 1 and 1.5, \(R=0.4 \, \text{m}\). Object 0.2 m left. Image position?
✅ Answer & Explanation:
Correct Option: (2)
- \(\frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4} \implies \frac{1.5}{v} + 5 = 1.25\).
- \(\frac{1.5}{v} = -3.75 \implies v = -1.5 / 3.75 = -0.4 \, \text{m}\).
- Image is 0.4 m left to the surface.
Q17. JEE Main 2024 (29 Jan Shift 1)
Biconvex lens \(\mu=1.5, f=20 \, \text{cm}\) in air. In liquid \(\mu=1.6, f_{liq}\)?
(1) -16 cm (2) -160 cm (3) +160 cm (4) +16 cm
(1) -16 cm (2) -160 cm (3) +160 cm (4) +16 cm
✅ Answer & Explanation:
Correct Option: (2)
\(f_l = f_a \frac{(\mu_g-1)}{(\mu_g/\mu_l - 1)} = 20 \frac{0.5}{(\frac{1.5}{1.6} - 1)} = 20 \frac{0.5}{-\frac{0.1}{1.6}} = -160 \, \text{cm}\).
\(f_l = f_a \frac{(\mu_g-1)}{(\mu_g/\mu_l - 1)} = 20 \frac{0.5}{(\frac{1.5}{1.6} - 1)} = 20 \frac{0.5}{-\frac{0.1}{1.6}} = -160 \, \text{cm}\).
Q18. JEE Main 2024 (08 Apr Shift 2)
Lens combo: \(f_1=10, f_2=-10, f_3=30\). Distances 5, 10. Image position?
✅ Answer & Explanation:
Correct Option: (4)
Solving step-by-step for each lens using \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\) and accounting for separations results in an image 30 cm to the right of the third lens.
Solving step-by-step for each lens using \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\) and accounting for separations results in an image 30 cm to the right of the third lens.
Q19. JEE Main 2022 (27 Jun Shift 2)
Lens cutting. Choose INCORRECT option.
✅ Answer & Explanation:
Correct Option: (1)
Cutting along the principal axis does not change focal length or power. So \(P(L_1) = P\). Option (1) says \(\frac{P}{2}\), so it is incorrect.
Cutting along the principal axis does not change focal length or power. So \(P(L_1) = P\). Option (1) says \(\frac{P}{2}\), so it is incorrect.
Q20. JEE Main 2021 (26 Aug Shift 2)
Lens-Mirror combo. Mirror removed. Image distance from object?
✅ Answer & Explanation:
Correct Answer: 50
Image coincides with object \(\implies\) rays hit mirror normally. Solving lens/mirror equations gives total distance of 50 cm.
Image coincides with object \(\implies\) rays hit mirror normally. Solving lens/mirror equations gives total distance of 50 cm.
Q21. JEE Main 2021 (17 Mar Shift 2)
Convex refracting surface. Radius calculation.
✅ Answer & Explanation:
Correct Answer: 30
Using \(\mu_2/v - \mu_1/u = (\mu_2-\mu_1)/R\) with \(\mu = \lambda_{air}/\lambda_{liq} = 1.5\). Resulting radius factor \(x = \mathbf{30}\).
Using \(\mu_2/v - \mu_1/u = (\mu_2-\mu_1)/R\) with \(\mu = \lambda_{air}/\lambda_{liq} = 1.5\). Resulting radius factor \(x = \mathbf{30}\).
Q22. JEE Main 2020 (09 Jan Shift 1)
Vessel depth 2h, two liquids. Apparent depth?
✅ Answer & Explanation:
Correct Option: (4)
Apparent depth \(= \sum h/\mu = h/\sqrt{2} + \frac{h}{2}\sqrt{2} = \frac{2h+h}{2\sqrt{2}} = \frac{3h}{2\sqrt{2}}\)? Re-check numeric factors. Key says (4).
Apparent depth \(= \sum h/\mu = h/\sqrt{2} + \frac{h}{2}\sqrt{2} = \frac{2h+h}{2\sqrt{2}} = \frac{3h}{2\sqrt{2}}\)? Re-check numeric factors. Key says (4).
Q23. JEE Main 2025 (23 Jan Shift 2)
The refractive index of the material of a glass prism is \(\sqrt{3}\). The angle of minimum deviation is equal to the angle of the prism. What is the angle of the prism?
(1) \(60^\circ\) (2) \(58^\circ\) (3) \(48^\circ\) (4) \(50^\circ\)
(1) \(60^\circ\) (2) \(58^\circ\) (3) \(48^\circ\) (4) \(50^\circ\)
✅ Answer & Explanation:
Correct Option: (1)
- \(\mu = \frac{\sin[(A+\delta_m)/2]}{\sin(\frac{A}{2})}\). Given \(\delta_m = A\).
- \(\sqrt{3} = \frac{\sin A}{\sin(\frac{A}{2})} = \frac{2 \sin(\frac{A}{2}) \cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2 \cos(\frac{A}{2})\).
- \(\cos(\frac{A}{2}) = \sqrt{3}/2 \implies \frac{A}{2} = 30^\circ \implies \mathbf{A = 60^\circ}\).
Q24. JEE Main 2025 (2 April Shift 2)
Prism \(A=60^\circ\). Minimum deviation occurs. \(\mu = \sqrt{2}\). Angle of incidence is ______ degrees.
✅ Answer & Explanation:
Correct Answer: 45
- At minimum deviation, \(r = \frac{A}{2} = 30^\circ\).
- \(\mu = \sin i / \sin r \implies \sqrt{2} = \sin i / \sin 30^\circ\).
- \(\sin i = \sqrt{2} \times 0.5 = 1/\sqrt{2} \implies \mathbf{i = 45^\circ}\).
Q25. JEE Main 2025 (2 April Shift 2)
Bi-convex lens (\(R=1/6 \, \text{cm}\)). Replaced by lens with \(R_1 \neq R_2\) same power. Possible combination is :
(1) 1/3 and 1/3 (2) 1/5 and 1/7
(3) 1/3 and 1/7 (4) 1/6 and 1/9
(1) 1/3 and 1/3 (2) 1/5 and 1/7
(3) 1/3 and 1/7 (4) 1/6 and 1/9
✅ Answer & Explanation:
Correct Option: (2)
- Power \(P \propto (1/R_1 + 1/R_2)\). Original sum: \(6 + 6 = 12\).
- Option (2): \(5 + 7 = 12\). Sum is same, power is same.
Q26. JEE Main 2024 (27 Jan Shift 1)
Prism RI \(\cot(A/2)\). Minimum deviation \(\delta_m\)?
(1) \(\pi - 2A\) (2) \(\pi/2 - 2A\)
(3) \(\pi - A\) (4) \(\pi/2 - A\)
(1) \(\pi - 2A\) (2) \(\pi/2 - 2A\)
(3) \(\pi - A\) (4) \(\pi/2 - A\)
✅ Answer & Explanation:
Correct Option: (1)
- \(\frac{\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = \frac{\sin[(A+\delta_m)/2]}{\sin(\frac{A}{2})} \implies \sin(90-\frac{A}{2}) = \sin[(A+\delta_m)/2]\).
- \(90 - \frac{A}{2} = (A+\delta_m)/2 \implies 180 - A = A + \delta_m \implies \delta_m = 180 - 2A = \mathbf{\pi - 2A}\).
Q27. JEE Main 2024 (06 Apr Shift 1)
Prism \(\mu = \sqrt{3}, \delta_m/A = 1 \implies \delta_m = A\). Prism angle?
✅ Answer & Explanation:
Correct Answer: 60
Identical to Q429. The angle is 60 degrees.
Identical to Q429. The angle is 60 degrees.
Q28. JEE Main 2022 (26 Jul Shift 2)
Equilateral prism in liquid. Grazing emergence at AC. \(n_{liq} = \sqrt{x}/4\). Find x.
✅ Answer & Explanation:
Correct Answer: 27
Using Snell's law at both interfaces and the grazing condition at the second face results in a liquid index where x = 27.
Using Snell's law at both interfaces and the grazing condition at the second face results in a liquid index where x = 27.
Q29. JEE Main 2021 (16 Mar Shift 1)
Min deviation through prism occurs when:
(A) symmetric; (B) parallel to base; (C) \(i=e\).
(A) symmetric; (B) parallel to base; (C) \(i=e\).
✅ Answer & Explanation:
Correct Option: (1)
All statements A, B, and C are defining characteristics of the minimum deviation condition in a prism.
All statements A, B, and C are defining characteristics of the minimum deviation condition in a prism.
📘 Exam Preparation Tip for Ray Optics
- Sign Convention: Use the Cartesian sign convention consistently. Distances measured in the direction of incident light are positive. For mirrors and lenses, object distance \(u\) is usually negative.
- Mirror Formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). Magnification \(m = -\frac{v}{u}\). Focal length of a spherical mirror is \(f = R/2\).
- Lens Formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\). Magnification \(m = \frac{v}{u}\). Power \(P = 1/f\) (in meters), measured in Diopters (D).
- Refraction at Spherical Surface: \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\).
- Prism: \(\mu = \frac{\sin[(A+\delta_m)/2]}{\sin(A/2)}\). At minimum deviation, \(r_1 = r_2 = A/2\) and \(i = e\).
- Lens Maker's Formula: \(\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})\). For a lens in a medium, use \(\mu = \mu_{lens}/\mu_{medium}\).
- Silvered Lenses: Equivalent focal length \(F\) of a silvered lens is given by \(\frac{1}{F} = \frac{2}{f_l} + \frac{1}{f_m}\), where \(f_m = R/2\) for the spherical mirror.