A small candle of height 2.5 cm is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed to obtain a sharp image? Describe the image. What happens as the candle is moved closer to the mirror?
\(R = -36\) cm ⇒ \(f = -18\) cm. \(u = -27\) cm.
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-18}-\frac{1}{-27} = \frac{-3+2}{54} = -\frac{1}{54}\) ⇒ \(v = -54\) cm.
Screen at 54 cm in front of mirror. \(m = -v/u = -54/27\cdot(-1) = -2\). \(h' = -5\) cm — real, inverted, magnified (twice).
As the candle is moved closer to the mirror (towards F), \(|v|\) increases and the image grows larger, moving away to infinity. If brought inside the focal length, a virtual enlarged image is formed behind the mirror and no screen will capture it.
Exercise 9.2
A 1.2 cm size object is placed 9 cm in front of a convex mirror of focal length 15 cm. Find the position, size and nature of the image. What happens as the object is moved farther away?
\(u=-9, f=+15\). \(\frac{1}{v}=\frac{1}{15}+\frac{1}{9}=\frac{3+5}{45}=\frac{8}{45}\) ⇒ \(v = +5.625\) cm (behind).
\(m = -v/u = -5.625/(-9) = +0.625\). \(h' = 0.75\) cm. Virtual, erect, diminished.
As object moves away, the image remains virtual and erect but shrinks further, tending toward the focus.
Exercise 9.3
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom is measured to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63, to what height will the microscope have to be moved to focus on the needle again?
\(n_w = h/h' = 12.5/9.4 = 1.33\).
For liquid (\(n=1.63\)): \(h' = 12.5/1.63 = 7.67\) cm. Microscope must be lowered by \(9.4 - 7.67 = 1.73\) cm.
Exercise 9.4
A ray of light incident on one face of a glass slab (\(n = 1.55\)) at 60° passes into the glass. Find the angle of refraction. Also find the speed of light in the glass.
A small bulb is placed at the bottom of a tank filled with water to a depth of 80 cm. What is the area of the surface through which light from the bulb can emerge out? Refractive index of water is 1.33.
Light escapes only within a cone of half-angle equal to the critical angle.
\(\sin\theta_c = 1/1.33 = 0.7519\), \(\theta_c = 48.75°\); \(\tan\theta_c = 1.137\).
Radius of the illuminated circle on the surface: \(r = h\tan\theta_c = 0.80\times 1.137 = 0.91\) m.
Area \(= \pi r^2 = 3.14\times 0.828 = \boxed{2.60\ \text{m}^2}\).
Exercise 9.6
A prism of angle 60° produces a minimum angle of deviation of 40°. Find its refractive index. If the prism is placed in water (\(n_w = 1.33\)), predict the new minimum deviation.
\(n = \sin[(60+40)/2]/\sin 30° = \sin 50°/\sin 30° = 0.766/0.5 = 1.532\).
In water: relative index \(n_{gw} = 1.532/1.33 = 1.152\). Solve \(\sin[(60+\delta_m')/2] = 1.152\times\sin 30° = 0.576\) ⇒ \((60+\delta_m')/2 = 35.17°\) ⇒ \(\delta_m' \approx 10.3°\). Minimum deviation is much smaller in water.
Exercise 9.7
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What radius is required if the focal length is to be 20 cm?
A beam of light converges to a point P. A lens is placed 12 cm in front of P. At what point does the beam converge if the lens is (a) convex of f = 20 cm (b) concave of f = 16 cm?
The converging beam is equivalent to a virtual object at P, 12 cm beyond the lens ⇒ \(u = +12\) cm.
(a) \(\frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3+5}{60} = \frac{8}{60}\), \(v = 7.5\) cm on the far side of the lens.
(b) \(\frac{1}{v} = \frac{1}{-16} + \frac{1}{12} = \frac{-3+4}{48} = \frac{1}{48}\), \(v = +48\) cm on the far side.
Exercise 9.9
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image.
\(u=-14, f=-21\). \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=-\frac{1}{21}-\frac{1}{14} = -\frac{5}{42}\) ⇒ \(v = -8.4\) cm.
\(m = v/u = -8.4/-14 = 0.6\); \(h' = 1.8\) cm. Virtual, erect, diminished, 8.4 cm in front of the lens.
Exercise 9.10
A convex lens of focal length 30 cm is in contact with a concave lens of focal length 20 cm. Find the power of the combination. Is the system converging or diverging? Ignore thickness.
\(P_1 = 100/30 = 3.33\) D; \(P_2 = -100/20 = -5\) D. \(P = -1.67\) D. Net diverging. \(f = 1/P = -60\) cm.
Exercise 9.11
A compound microscope has objective focal length 2.0 cm and eyepiece focal length 6.25 cm and separation 15 cm. Find the distance of the object for the final image to form at (a) the near point (25 cm) (b) infinity. Also calculate the magnifying power in each case.
(a) For eyepiece: \(v_e = -25, f_e = 6.25\). \(\frac{1}{u_e}=\frac{1}{v_e}-\frac{1}{f_e}=-\frac{1}{25}-\frac{1}{6.25} = -\frac{1+4}{25}=-\frac{5}{25}\) ⇒ \(u_e = -5\) cm. So the intermediate image is 5 cm in front of the eyepiece, i.e. 10 cm from objective. For objective: \(v_o = +10\). \(\frac{1}{u_o}=\frac{1}{10}-\frac{1}{2}=-\frac{4}{10}\) ⇒ \(u_o = -2.5\) cm. \(m = (v_o/u_o)(1+D/f_e) = (-10/2.5)\times(1+4) = -20\).
(b) For infinity: \(u_e = -f_e = -6.25\), so \(v_o = 15 - 6.25 = 8.75\) cm. \(\frac{1}{u_o}=\frac{1}{8.75}-\frac{1}{2} = -\frac{6.75}{17.5}\) ⇒ \(u_o = -2.59\) cm. \(m = (v_o/u_o)\cdot(D/f_e) = -3.38\times 4 = -13.5\).
Exercise 9.12
A person with a normal near point (25 cm) using a compound microscope with objective f_o = 8 mm and eyepiece f_e = 2.5 cm can bring an object 9.0 mm from the objective into sharp focus. What is the separation between the two lenses? How much is the magnifying power of the microscope?
Objective: \(u_o=-0.9\) cm, \(f_o=0.8\) cm. \(\frac{1}{v_o}=\frac{1}{0.8}-\frac{1}{0.9}=\frac{1}{7.2}\) ⇒ \(v_o=7.2\) cm.
Eyepiece (final at near point): \(v_e=-25, f_e=2.5\). \(\frac{1}{u_e}=-\frac{1}{25}-\frac{1}{2.5}=-\frac{11}{25}\) ⇒ \(u_e=-2.27\) cm.
Separation \(= v_o + |u_e| = 7.2 + 2.27 = \boxed{9.47\ \text{cm}}\).
\(m = (v_o/|u_o|)(1+D/f_e) = (7.2/0.9)(1+10) = 8\times 11 = 88\).
Exercise 9.13
A small telescope has objective of focal length 144 cm and eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the two lenses? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image?
\(m = 144/6 = \boxed{24}\); \(L = 150\) cm.
Angular size of tower = 100/3000 = 1/30 rad. Intermediate image height at objective focal plane = \(f_o \tan\alpha \approx 144/30 = 4.8\) cm. Final angular size \(= m\times\alpha\). Linear size at 25 cm \(\approx 25\times (1/30)\times 24 = 20\) cm.
Exercise 9.14
(a) A giant refracting telescope has an objective of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification? (b) If the telescope is used to view the moon (diameter 3.48×10⁶ m, distance 3.8×10⁸ m), what is the diameter of the moon's image formed by the objective?
(a) \(m = 1500/1 = \boxed{1500}\).
(b) Angular size of moon = \(3.48\times 10^6/3.8\times 10^8 = 9.16\times 10^{-3}\) rad. Diameter of image = \(f_o\cdot\alpha = 15\times 9.16\times 10^{-3} = 0.1374\) m = 13.74 cm.
Exercise 9.15
Use the mirror formula to deduce that (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f; (b) a convex mirror always produces a virtual image of any object; (c) a concave mirror with object between pole and focus produces a virtual magnified image.
(a) For concave, \(f<0\). If \(2f < u < f\) (i.e. \(|u|\) between \(|f|\) and \(2|f|\)), then \(1/v = 1/f - 1/u\) yields \(v < 2f\) (beyond 2f, real).
(b) Convex: \(f>0, u<0\). \(1/v = 1/f - 1/u\) → always positive → \(v>0\): virtual image behind mirror.
(c) Concave, \(f0\): virtual. Also \(|v|>|u|\) ⇒ \(|m|>1\) magnified.
Exercise 9.16
A small pin fixed on a table top is viewed from above a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed through a glass slab 15 cm thick and refractive index 1.5? Does the answer depend on the observer's location?
Apparent shift = \(t(1 - 1/n) = 15(1 - 1/1.5) = 15\times(1/3) = \boxed{5\ \text{cm}}\).
For near-normal viewing the shift is independent of the observer's distance, but for strongly oblique viewing it does change (paraxial assumption breaks down).
Exercise 9.17
(a) Figure shows a cross-section of a light pipe made of glass (\(n=1.68\)) and an outer coating of refractive index 1.44. What is the range of angles of incident rays with the axis of the pipe for which total internal reflections inside the pipe occur? (b) What is the answer if there is no outer coating?
(a) Critical angle at core–cladding: \(\sin\theta_c = 1.44/1.68 = 0.857\), \(\theta_c = 59°\). Inside the pipe, the ray must hit the wall at ≥ 59° from the normal, i.e. at ≤ 31° from the axis. Applying Snell at the entrance face: \(\sin i_{max} = 1.68\sin 31° = 0.866\), \(i_{max} = 60°\). Range: 0° to 60°.
(b) Without cladding, critical angle between glass and air = \(\sin^{-1}(1/1.68) = 37°\). Ray must hit wall at ≥ 37° from normal, i.e. ≤ 53° from axis — every ray entering at any angle with the axis undergoes TIR (all angles 0°–90° work).
Exercise 9.18
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?
For a fixed object–image distance \(D\), real image forms only if \(D \geq 4f\), with equality at unit magnification (object and image both at 2f).
\(f_{max} = D/4 = 300/4 = \boxed{75\ \text{cm}}\).
Exercise 9.19
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different positions separated by 20 cm. Determine the focal length of the lens.
(a) Determine the effective focal length of the combination of two lenses of focal lengths +30 cm and –20 cm placed 8 cm apart. (b) Obtain the magnification produced by an object 1.5 cm tall placed 40 cm in front of the convex lens.
(a) \(\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = \frac{1}{30} - \frac{1}{20} - \frac{8}{30\times(-20)} = \frac{2-3}{60} + \frac{8}{600} = -\frac{10}{600}+\frac{8}{600}=-\frac{2}{600}\). \(f = -300\) cm (diverging).
(b) Convex lens first: \(u=-40, f_1=+30\). \(\frac{1}{v}=1/30-1/40 = 1/120\), \(v = +120\) cm, \(m_1 = 120/(-40) = -3\). Intermediate image is 120 cm past first lens, i.e. 112 cm past the concave lens (so acts as virtual object at \(u_2 = +112\)). For concave \(f_2=-20\): \(1/v_2 = 1/(-20)+1/112\) — gives \(v_2 \approx -24.4\) cm. \(m_2 = v_2/u_2 = -0.218\). Net \(m = m_1 m_2 \approx +0.65\). Image height \(\approx 1.5 \times 0.65 = 0.98\) cm.
Exercise 9.21
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? Refractive index of the prism material is 1.524.
Critical angle: \(\sin\theta_c = 1/1.524 = 0.6562\), \(\theta_c = 41°\).
At the second face, for TIR just to occur, \(r_2 = \theta_c = 41°\). Thus \(r_1 = A - r_2 = 60 - 41 = 19°\).
At first face: \(\sin i = n\sin r_1 = 1.524\times\sin 19° = 1.524\times 0.3256 = 0.496\). \(\boxed{i \approx 29.75°}\).
Exercise 9.22
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (converging lens of focal length 10 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification of the magnifier?
(a) \(u=-9, f=+10\). \(1/v = 1/10 - 1/9 = -1/90\), \(v = -90\) cm. \(m = v/u = -90/-9 = 10\). Each side becomes 10 mm ⇒ area = \(\boxed{100\ \text{mm}^2}\).
(b) Angular magnification \(= D/u = 25/9 \approx 2.78\).
Exercise 9.23
(a) At what distance should the card sheet in Exercise 9.22 be placed from the lens to get the maximum possible angular magnification? (b) What is the maximum angular magnification obtainable?
(a) Maximum angular magnification occurs with image at near point (\(v = -25\) cm). \(\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = -1/25 - 1/10 = -7/50\) ⇒ \(u \approx -7.14\) cm.
(b) \(m_{max} = 1 + D/f = 1 + 25/10 = \boxed{3.5}\).
Exercise 9.24
An angular magnification of 30× is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Final image at near point: \(m_e = 1 + 25/5 = 6\). So \(m_o = 30/6 = 5\). \(m_o = v_o/u_o = 5\) ⇒ \(v_o = 5|u_o|\). From objective formula with \(f_o = 1.25\): \(1/v_o - 1/u_o = 1/1.25\) → \(1/(5|u_o|) + 1/|u_o| = 0.8\) → \(6/(5|u_o|) = 0.8\) → \(|u_o| = 1.5\) cm. \(v_o = 7.5\) cm.
For eyepiece: \(v_e = -25, f_e = 5\): \(1/u_e = -1/25 - 1/5 = -6/25\), \(u_e = -25/6 = -4.17\) cm. Separation \(L = v_o + |u_e| = 7.5 + 4.17 = \boxed{11.67\ \text{cm}}\).
Exercise 9.25
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment, (b) the final image is formed at the least distance of distinct vision (25 cm).
Frequently Asked Questions - NCERT Exercises and Solutions: Ray Optics and Optical Instruments
What are the key NCERT exercise types in Chapter 9 Ray Optics and Optical Instruments?
NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Ray Optics and Optical Instruments?
For numerical problems in NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 9?
From NCERT Class 12 Physics Chapter 9 (Ray Optics and Optical Instruments), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 9 Ray Optics and Optical Instruments problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 9 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
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MyAiSchool Class 12 Physics Chapter 9 Ray Optics and Optical Instruments solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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