Ray Optics

Answer: (b)

1. A parallel beam meets at the principal focus, which lies midway between the pole and the centre of curvature.
2. $F = \frac{R}{2} \Rightarrow R = 2F$
3. So the answer is (b) $R = 2F$.

Answer: (c)

1. The parallax method locates a real, inverted image with a pin, so the image must be real.
2. A concave mirror forms a real image for every object position beyond the focus $F$.
3. So the answer is (c) at any point beyond $F$.

Answer: 10

1. $m = \frac{f}{f - u}$ with $f = -10\text{ cm}$.
2. Real image, $m = -2$: $f - u = -\frac{f}{2} \Rightarrow u = \frac{3f}{2} = -15\text{ cm}$
3. Virtual image, $m = +2$: $f - u = \frac{f}{2} \Rightarrow u = \frac{f}{2} = -5\text{ cm}$
4. Distance between the two positions $= 15 - 5 = 10\text{ cm}$

Answer: (d)

1. Real image three times magnified: $m = -3 \Rightarrow v = 3u$ (both in front of the mirror).
2. Object-image distance $= |v - u| = 2|u| = 40 \Rightarrow u = -20\text{ cm}$, $v = -60\text{ cm}$
3. $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = -\frac{1}{60} - \frac{1}{20} = -\frac{1}{15}$
4. $f = -15\text{ cm}$, so the answer is (d).

Answer: (b)

1. $f = -10\text{ cm}$, so $C$ is at 20 cm. The near end at $u = -20$ cm images at $v = -20$ cm.
2. Far end at $u = -30$ cm: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{10} + \frac{1}{30} = -\frac{1}{15} \Rightarrow v = -15\text{ cm}$
3. Image length $= 20 - 15 = 5\text{ cm}$
4. So the answer is (b) 5.

Answer: (b)

1. For a mirror $f = \frac{R}{2}$, which depends only on the geometry of the surface.
2. Reflection does not depend on the refractive index of the surrounding medium, so $f$ is unchanged.
3. So the answer is (b) $f$.

Answer: (b)

1. $|f| = 6\text{ cm}$, $u = -18\text{ cm}$, $m = \frac{f}{f - u}$
2. Convex ($f = +6$): $m = \frac{6}{6 + 18} = \frac{1}{4}$
3. Concave ($f = -6$): $m = \frac{-6}{-6 + 18} = -\frac{1}{2}$
4. Ratio of sizes $= \frac{1/4}{1/2} = \frac{1}{2}$, so the answer is (b).

Answer: 45

1. $m = -\frac{v}{u} = -\frac{1}{3} \Rightarrow v = \frac{u}{3}$ (real image, concave mirror).
2. Object-image distance $= |u| - \frac{|u|}{3} = \frac{2|u|}{3} = 30 \Rightarrow u = -45\text{ cm}$, $v = -15\text{ cm}$
3. $\frac{1}{f} = -\frac{1}{15} - \frac{1}{45} = -\frac{4}{45} \Rightarrow f = -\frac{45}{4}\text{ cm}$
4. $|x| = 45$

Answer: 8

1. Convex mirror: $f = +1\text{ m}$; $u = -24\text{ m}$; $\frac{du}{dt} = +25\text{ m/s}$ (90 km/h), constant.
2. $v = \frac{fu}{u - f} \Rightarrow \frac{dv}{dt} = -\frac{f^2}{(u - f)^2}\frac{du}{dt}$
3. $a = \frac{d^2v}{dt^2} = \frac{2f^2}{(u - f)^3}\left(\frac{du}{dt}\right)^2 = \frac{2(1)(625)}{(-25)^3}$
4. $|a| = \frac{1250}{15625} = 0.08\text{ m/s}^2 \Rightarrow 100a = 8\text{ m/s}^2$

Answer: (c)

1. Erect diminished image ($m = \frac{1}{4}$) means a convex mirror: $u = -x$, $v = +\frac{x}{4}$.
2. Object-image distance: $x + \frac{x}{4} = 40 \Rightarrow x = 32\text{ cm}$, $v = 8\text{ cm}$
3. $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{8} - \frac{1}{32} = \frac{3}{32} \Rightarrow f = 10.67\text{ cm}$
4. (Taking $m = -\frac{1}{4}$ with a concave mirror also gives $|f| = 10.7$ cm.) So the answer is (c) 10.7 cm.

Answer: (d)

1. Erect diminished image, so convex mirror. Use $m = \frac{f}{f-u}$.
2. $u = -40$: $\frac{f}{f+40} = \frac{1}{2} \Rightarrow f = 40\text{ cm}$
3. For $m = \frac{1}{3}$: $\frac{40}{40 + x} = \frac{1}{3} \Rightarrow x = 80\text{ cm}$
4. Object moves from 40 cm to 80 cm. So the answer is (d) 40 cm away from the mirror.

Answer: (b)

1. $m = -3 \Rightarrow |v| = 3|u|$, both real (in front).
2. $|v| - |u| = 2|u| = 20 \Rightarrow |u| = 10\text{ cm}$, $|v| = 30\text{ cm}$
3. $\frac{1}{f} = \frac{1}{10} + \frac{1}{30} = \frac{4}{30} \Rightarrow f = 7.5\text{ cm}$
4. $R = 2f = 15\text{ cm}$. So the answer is (b) 15 cm.

Answer: (d)

1. Convex mirror: $f = +10\text{ m}$. Use $\frac{1}{u} = \frac{1}{f} - \frac{1}{v}$.
2. $v = \frac{25}{3}$: $\frac{1}{u} = \frac{1}{10} - \frac{3}{25} = -\frac{1}{50} \Rightarrow u = -50\text{ m}$
3. $v = \frac{50}{7}$: $\frac{1}{u} = \frac{1}{10} - \frac{7}{50} = -\frac{1}{25} \Rightarrow u = -25\text{ m}$
4. Speed $= \frac{25}{30} = \frac{5}{6}\text{ m/s} = \frac{5}{6} \times \frac{18}{5} = 3\text{ km/hr}$
5. So the answer is (d) 3.

Answer: (d)

1. Convex mirror, $f = +30$, $u = -30$: $\frac{1}{v} = \frac{1}{30} + \frac{1}{30} \Rightarrow v = +15\text{ cm}$ (behind the mirror).
2. Object to this image: $30 + 15 = 45\text{ cm}$.
3. A plane mirror forms its image as far behind as the object is in front, so it must sit midway: $22.5$ cm from the object.
4. Distance between mirrors $= 30 - 22.5 = 7.5\text{ cm}$. So the answer is (d) 7.5 cm.

Answer: (c)

1. A plane mirror forms an erect image of the same size as the object.
2. $m = \frac{h_i}{h_o} = +1$
3. So the answer is (c) $+1$.

Answer: (c)

1. Statement I: an image forms only when rays are reflected or refracted in a regular way so they meet (or appear to meet) at a point. Correct.
2. Statement II: an object looks coloured because it reflects some of the colours present in the incident light. Correct.
3. So the answer is (c) both statements are correct.

Answer: (b)

1. Convex mirror: $f = +\frac{R}{2} = +15\text{ cm}$; image is virtual and erect, $m = +\frac{1}{2}$.
2. $m = -\frac{v}{u} = \frac{1}{2} \Rightarrow v = -\frac{u}{2}$
3. $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}: \ -\frac{2}{u} + \frac{1}{u} = -\frac{1}{u} = \frac{1}{15}$
4. $u = -15\text{ cm}$, so the answer is (b) $-15$ cm.

Answer: (c)

1. Real image: $m = -n = -\frac{v}{u} \Rightarrow v = nu$ (both negative).
2. Put $u = -x$, $v = -nx$, $f \to -f$ in $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$: $\frac{1}{nx} + \frac{1}{x} = \frac{1}{f}$
3. $\frac{n+1}{nx} = \frac{1}{f} \Rightarrow x = \left(\frac{n+1}{n}\right)f$
4. So the answer is (c) $\left(\frac{n+1}{n}\right)f$.

Answer: (a)

1. Magnified virtual image: concave mirror, $u = -x$, $v = +2x$.
2. Object-image distance $x + 2x = 15 \Rightarrow x = 5\text{ cm}$, $v = +10\text{ cm}$
3. $\frac{1}{f} = \frac{1}{10} - \frac{1}{5} = -\frac{1}{10} \Rightarrow f = -10\text{ cm}$
4. So the answer is (a) $-10$ cm.

Answer: (c)

1. Angle of incidence = angle of reflection $= 30^\circ$.
2. Deviation $\delta = 180^\circ - 2i = 180^\circ - 60^\circ = 120^\circ$
3. So the answer is (c) $120^\circ$.

Answer: (a)

1. A plane mirror image is virtual (A false), erect (B true), the same size (C false) and laterally inverted (D true).
2. So the answer is (a) B and D only.

Answer: (d)

1. Number of images $n = \frac{360^\circ}{\theta} - 1$ (when $\frac{360^\circ}{\theta}$ is even).
2. $3 = \frac{360^\circ}{\theta} - 1 \Rightarrow \theta = 90^\circ$
3. So the answer is (d) $90^\circ$.

Answer: (d)

1. Before: object 12 cm from the mirror, image 12 cm behind it, i.e. 24 cm from the object.
2. After moving the mirror 4 cm closer: object distance 8 cm, image 8 cm behind it, i.e. 16 cm from the object.
3. The image moves $24 - 16 = 8\text{ cm}$ (twice the mirror's shift), in the same direction as the mirror.
4. So the answer is (d) 8 cm towards the mirror.

Answer: (c)

1. $f = -20\text{ cm}$.
2. A ($u = -15$, inside F): $\frac{1}{v} = -\frac{1}{20} + \frac{1}{15} = \frac{1}{60} \Rightarrow v = +60\text{ cm}$ (behind the mirror)
3. B ($u = -25$): $\frac{1}{v} = -\frac{1}{20} + \frac{1}{25} = -\frac{1}{100} \Rightarrow v = -100\text{ cm}$ (in front)
4. Distance between images $= 60 + 100 = 160\text{ cm}$, so the answer is (c).

Answer: 18

1. Place A at $x = 0$, B at $x = 10$, O at $x = 2$.
2. Nearest image behind A: image of O in A at $x = -2$.
3. Image of O in B is at $x = 18$; its image in A is at $x = -18$.
4. Second nearest image behind A is $18\text{ cm}$ from A.

Answer: 32

1. $f = -20\text{ cm}$; the rod ends are at $u = -35\text{ cm}$ and $u = -45\text{ cm}$.
2. $u = -35$: $\frac{1}{v} = -\frac{1}{20} + \frac{1}{35} = -\frac{3}{140} \Rightarrow v = -\frac{140}{3}\text{ cm}$
3. $u = -45$: $\frac{1}{v} = -\frac{1}{20} + \frac{1}{45} = -\frac{1}{36} \Rightarrow v = -36\text{ cm}$
4. Image length $= \frac{140}{3} - 36 = \frac{32}{3}\text{ cm} \Rightarrow x = 32$

Answer: 32

1. Mirror formula (magnitudes, real image): $\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{40} + \frac{1}{120} = \frac{4}{120} \Rightarrow f = 30\text{ cm}$
2. Least count of the scale: $\Delta u = \Delta v = \frac{1}{20} = 0.05\text{ cm}$
3. Differentiating: $\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}$
4. $\Delta f = 900\left(\frac{0.05}{1600} + \frac{0.05}{14400}\right) = 900 \times \frac{0.5}{14400} = \frac{1}{32}\text{ cm}$
5. So $K = 32$.

Answer: 210

1. At M2 the angle of reflection is $30^\circ$, so the ray meets M2 at $60^\circ$ to the surface; deviation at M2 $= 180^\circ - 2(30^\circ) = 120^\circ$.
2. Triangle of the mirrors and the ray: angle with M1 $= 180^\circ - 75^\circ - 60^\circ = 45^\circ$, so $\theta_1 = 45^\circ$.
3. Deviation at M1 $= 180^\circ - 2(45^\circ) = 90^\circ$
4. Total deviation $= 90^\circ + 120^\circ = 210^\circ$

Answer: 400

1. $R = 200\text{ cm} \Rightarrow f = -100\text{ cm}$. After 10 s the object has moved 20 cm: $u = -80\text{ cm}$.
2. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{100} + \frac{1}{80} = \frac{1}{400}$
3. $v = +400\text{ cm}$ (virtual image behind the mirror)
4. So the image is at 400 cm from the mirror.

Answer: (c)

1. For any spherical mirror the focus is midway between the pole and the centre of curvature.
2. For a convex mirror both $f$ and $r$ are positive (behind the mirror), so $f = +\frac{1}{2}r$.
3. So the answer is (c).

Answer: (a)

1. Object beyond C: $|u| = R + d_1$; image is between F and C: $|v| = R - d_2$.
2. $\frac{1}{R + d_1} + \frac{1}{R - d_2} = \frac{2}{R}$
3. $R(2R + d_1 - d_2) = 2(R + d_1)(R - d_2) \Rightarrow R(d_1 - d_2) = 2d_1d_2$
4. $R = \frac{2d_1d_2}{d_1 - d_2}$, so the answer is (a).

Answer: (b)

1. On reflection, the component of $\vec a$ along the normal $\vec c$ reverses; the component along the mirror is unchanged.
2. Normal component $= (\vec a \cdot \vec c)\vec c$, so $\vec b = \vec a - 2(\vec a \cdot \vec c)\vec c$
3. So the answer is (b).

Answer: (b)

1. Take the corner as origin, M1 along the y-axis, M2 along the x-axis, so $P = (a, 2a)$.
2. Image in M1: $I_1 = (-a, 2a)$; image in M2: $I_2 = (a, -2a)$; third image $I_3 = (-a, -2a)$.
3. $I_1I_2 = \sqrt{(2a)^2 + (4a)^2} = 2\sqrt{5}\,a = 4.6a$
4. So the answer is (b) $4.6a$.

Answer: (d)

1. The rear-view mirror is convex: $f = +0.1\text{ m}$, $u = -1.9\text{ m}$.
2. $m = \frac{f}{f - u} = \frac{0.1}{0.1 + 1.9} = \frac{1}{20}$
3. Image speed $v_I = m^2 v_O = \frac{1}{400} \times 40 = 0.1\text{ m/s}$
4. So the answer is (d) 0.1 m s$^{-1}$.

Answer: (b)

1. Image is erect (same orientation) and diminished: $m = +\frac{25}{100} = +\frac{1}{4}$.
2. An erect, diminished image is formed only by a convex mirror, and an erect image from a single mirror is always virtual.
3. A virtual image of a mirror lies behind it, i.e. on the opposite side to the object.
4. So the answer is (b).

Answer: 150

1. Image $S'$ is 60 cm behind the mirror; rays from $S'$ through the mirror edges reach the man's line.
2. Distance of the man's line from $S'$: $60 + 120 = 180\text{ cm}$ = 3 times the mirror's distance from $S'$.
3. Visible stretch $= 50 \times \frac{180}{60} = 150\text{ cm}$
4. So the answer is 150 cm.

Answer: (b)

1. Concave mirror: $f \to -f$, $u = -1.5f$.
2. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{f} + \frac{1}{1.5f} = -\frac{1}{3f}$
3. $v = -3f$ (real image in front of the mirror).
4. So the answer is (b) $-3f$.

Answer: (a)

1. The hatching is on the outside, so the inner surface reflects: a concave mirror whose radius is the sphere's radius.
2. From the figure $R = 10\text{ cm}$, and the object is 10 cm from the pole, i.e. at the centre of curvature C.
3. An object at C gives an image at C: $v = u$, $m = -1$.
4. So the answer is (a) inverted, real and unmagnified.

Answer: (c)

1. Concave mirror, object at distance $x > f$: $|m| = \frac{f}{x - f}$
2. Just beyond F, $|m| \to \infty$; at $x = 2f$, $|m| = 1$; as $x \to \infty$, $|m| \to 0$.
3. The graph starts very high just after $f$, passes through $m = 1$ at $2f$ and keeps falling.
4. So the answer is (c).

Answer: 1

1. For a mirror, image speed along the axis: $v_I = \frac{v^2}{u^2}\, v_O$ (in magnitude).
2. $v_I = \left(\frac{10}{30}\right)^2 \times 9 = \frac{1}{9} \times 9$
3. So the image moves at $1\text{ cm s}^{-1}$.

Answer: (b)

1. Concave mirror: $f = -0.4\text{ m}$; upright image, $m = +5$.
2. $m = \frac{f}{f - u} \Rightarrow 5 = \frac{-0.4}{-0.4 - u}$
3. $-2 - 5u = -0.4 \Rightarrow u = -0.32\text{ m}$
4. So the answer is (b) 0.32 m.

Answer: (b)

1. Let the angle between the mirrors be $\theta$. The incoming ray is parallel to M2, so it makes $\theta$ with M1 and leaves M1 also at $\theta$.
2. In the triangle formed by the mirrors and the ray, the ray meets M2 at $180^\circ - 2\theta$ and leaves at the same angle.
3. Parallel to M1 means this angle equals $\theta$: $180^\circ - 2\theta = \theta$
4. $\theta = 60^\circ$, so the answer is (b).

Answer: (c)

1. The image $S'$ is a distance $L$ behind the mirror. Reflected rays seem to come from $S'$ and pass through the mirror's edges.
2. The man's line is $L + 2L = 3L$ from $S'$, while the mirror is $L$ from $S'$.
3. By similar triangles the visible stretch $= d \times \frac{3L}{L} = 3d$
4. So the answer is (c) $3d$.

Answer: (b)

1. $f = -15\text{ cm}$. For $u = -40$: $\frac{1}{v} = -\frac{1}{15} + \frac{1}{40} = -\frac{1}{24} \Rightarrow v = -24\text{ cm}$
2. For $u = -20$: $\frac{1}{v} = -\frac{1}{15} + \frac{1}{20} = -\frac{1}{60} \Rightarrow v = -60\text{ cm}$
3. The image shifts from 24 cm to 60 cm: $60 - 24 = 36\text{ cm}$ away from the mirror.
4. So the answer is (b).

Answer: (b)

1. $f = -5\text{ cm}$; object moves between $u = -8$ and $u = -12$ cm periodically, so the image is also periodic: (A) true, (B) false.
2. $v = \frac{uf}{u - f}$: $u = -8 \Rightarrow v = -\frac{40}{3} = -13.33$; $u = -10 \Rightarrow v = -10$; $u = -12 \Rightarrow v = -\frac{60}{7} = -8.57$ cm
3. Turning points are 3.33 cm and 1.43 cm from the image of $x = 10$ cm, so they are asymmetric: (C) true.
4. Distance $= \frac{40}{3} - \frac{60}{7} = \frac{100}{21}\text{ cm}$: (D) true.
5. So the answer is (b) (A), (C), (D).

Answer: (d)

1. When a mirror rotates by $\theta$, the reflected ray rotates by $2\theta$.
2. For a small angle, spot shift $y = x \cdot 2\theta$
3. $\theta = \frac{y}{2x}$, so the answer is (d).

Answer: (c)

1. Negative $m$: inverted, real image, formed only by a concave mirror. So A ($-2$) and B ($-\frac{1}{2}$) $\to$ q, r.
2. $m = +2$: erect and magnified, only a concave mirror (object inside $F$), virtual. C $\to$ q, s.
3. $m = +\frac{1}{2}$: erect and diminished, convex mirror, virtual. D $\to$ p, s.
4. So the answer is (c).

Answer: (c)

1. $u = -10\text{ cm}$. The virtual image is behind the mirror and 25 cm from the face, so $v = +15\text{ cm}$.
2. $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30} \Rightarrow f = -30\text{ cm}$
3. $|R| = 2|f| = 60\text{ cm}$ (concave mirror).
4. So the answer is (c) 60 cm.

Answer: (d)

1. Incidence $\theta$ at the first mirror, so the reflected ray makes $90^\circ - \theta$ with that mirror.
2. In the triangle formed by the two mirrors and the ray, the ray makes $180^\circ - 70^\circ - (90^\circ - \theta) = 20^\circ + \theta$ with the second mirror.
3. It leaves the second mirror at the same angle. To be parallel to the first mirror this angle must equal $70^\circ$.
4. $20^\circ + \theta = 70^\circ \Rightarrow \theta = 50^\circ$, so the answer is (d).

Answer: (d)

1. $f = -10\text{ cm}$. Near end at $u = -20 = 2f$, i.e. at C, so its image is also at 20 cm.
2. Far end at $u = -30$: $\frac{1}{v} = -\frac{1}{10} + \frac{1}{30} = -\frac{1}{15} \Rightarrow v = -15\text{ cm}$
3. Image length $= 20 - 15 = 5\text{ cm}$
4. So the answer is (d) 5 cm.

Answer: (a)

1. Convex mirror: $f = +0.2\text{ m}$, $u = -2.8\text{ m}$.
2. $m = \frac{f}{f - u} = \frac{0.2}{0.2 + 2.8} = \frac{1}{15}$
3. $v_I = m^2 v_O = \frac{1}{225} \times 15 = \frac{1}{15}\text{ m/s}$
4. So the answer is (a).

Answer: (b)

1. $n = \frac{360^\circ}{\theta} - 1$
2. $3 = \frac{360^\circ}{\theta} - 1 \Rightarrow \frac{360^\circ}{\theta} = 4 \Rightarrow \theta = 90^\circ$
3. So the answer is (b) $90^\circ$.

Answer: (a)

1. $\frac{360^\circ}{60^\circ} = 6$ (even), so $n = 6 - 1$.
2. $n = 5$
3. So the answer is (a) 5.

Answer: (b)

1. To see the full image, a plane mirror needs only half the height of the person.
2. $L_{min} = \frac{H}{2} = \frac{6}{2} = 3\text{ ft}$
3. So the answer is (b) 3 feet.

Answer: (b)

1. $x = \frac{n_w}{n_a}$, $y = \frac{n_g}{n_w}$, $z = \frac{n_a}{n_g}$
2. $xyz = \frac{n_w}{n_a} \cdot \frac{n_g}{n_w} \cdot \frac{n_a}{n_g} = 1$
3. So the answer is (b) $xyz = 1$.

Answer: (d)

1. Snell's law: $n = \frac{\sin i}{\sin r}$; for the same $i$, smallest $r$ gives largest $n$.
2. Speed $v = \frac{c}{n}$ is minimum where $n$ is maximum, i.e. $r = 20^\circ$.
3. So the answer is (d) D.