The image of an object is formed by rays from every point on the object reflecting off *all* parts of the mirror surface and converging (or appearing to diverge) to form the corresponding image point. If a portion of the mirror is covered, it simply means that the rays hitting that covered portion are no longer reflected to form the image. However, the remaining uncovered portion of the mirror can still reflect rays from the object to form the *entire* image.

Covering the lower half of the mirror will **not** result in only seeing the upper half of the object or half of the image. The **complete image** of the object will still be formed by the rays reflecting from the upper, uncovered half of the mirror.

The effect will be that the **intensity** (brightness) of the image will be reduced. This is because the total number of light rays from the object reaching the image point after reflection is reduced (by approximately half, since half the reflecting area is covered). This will make the image fainter.

The mobile phone is an extended object placed along the principal axis. Different points on the phone are at different distances from the mirror. Let's consider two points on the phone: one closer to the mirror (say, point B) and one farther from the mirror (say, point A). Point C is on the principal axis. An object along the principal axis is sometimes called a longitudinal object. The formula $m = -v/u$ for magnification applies to the lateral magnification of an object perpendicular to the principal axis. Magnification along the axis is different.

Diagram for image formation:

Let's consider the image of point B at distance $u_B$ and point A at distance $u_A$. The image positions $v_B$ and $v_A$ are given by the mirror equation: $1/v_B + 1/u_B = 1/f$ and $1/v_A + 1/u_A = 1/f$. Since the points A and B are at different distances ($u_A \ne u_B$), their images will be formed at different distances ($v_A \ne v_B$), and the magnification $m=-v/u$ will also be different for different parts of the object.

Why magnification is not uniform: The lateral magnification $m = -v/u$ depends on the object distance $u$. Since different parts of the mobile phone (along the principal axis) are at different distances from the mirror, the lateral magnification will be different for each part. This means that a part of the object which is, for example, $1$ cm wide (perpendicular to the axis) at one location will have a different width in the image compared to a part which is $1$ cm wide at a different location along the phone's length. This varying magnification along the length of the object causes the **distortion** of the image.

Will the distortion depend on location? Yes. The magnification $m = -v/u$ changes with $u$. The relationship between $u$ and $v$ is non-linear ($1/v = 1/f - 1/u$). The rate at which magnification changes along the object's length depends on the absolute position of the object relative to the focal length and radius of curvature of the mirror. For example, if the entire phone is far from the mirror compared to its focal length, the object distances for different parts of the phone are relatively similar, and the distortion might be less noticeable. If the phone spans across the focal point or is very close to the mirror, the object distances vary significantly in terms of their relation to $f$, leading to more pronounced variations in magnification and thus greater distortion. The distortion of the image will definitely depend on the location of the phone with respect to the mirror.

Given: Concave mirror, radius of curvature $R = 15$ cm. For a concave mirror, $R$ is taken as negative according to sign convention, so $R = -15$ cm. Focal length $f = R/2 = -15/2 = -7.5$ cm. The object is placed in front of the mirror, so object distance $u$ is negative.

We use the mirror equation $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ and magnification formula $m = -v/u$.

(i) Object placed at 10 cm in front: Object distance $u = -10$ cm.

$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5}$.

$\frac{1}{v} = \frac{1}{10} - \frac{1}{7.5} = \frac{1}{10} - \frac{1}{15/2} = \frac{1}{10} - \frac{2}{15}$.

Find a common denominator (30): $\frac{1}{v} = \frac{3}{30} - \frac{4}{30} = \frac{-1}{30}$.

$v = -30$ cm.

Image position: The image is formed at 30 cm in front of the mirror (since $v$ is negative). This is beyond the center of curvature ($R=-15$ cm, so 2f=-15 cm). The object is between f (-7.5 cm) and 2f (-15 cm). Expected image is real, inverted, and magnified, beyond 2f.

Magnification $m = -v/u = -(-30 \text{ cm}) / (-10 \text{ cm}) = 30 / (-10) = -3$.

Nature of image: Since $v$ is negative, the image is formed on the same side as the object, so it is a **real image**. Since $m$ is negative, the image is **inverted**. Since $|m| = 3 > 1$, the image is **magnified**.

Position: 30 cm in front of the mirror. Nature: Real, inverted, magnified. Magnification: -3.

(ii) Object placed at 5 cm in front: Object distance $u = -5$ cm.

$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5}$.

$\frac{1}{v} = \frac{1}{5} - \frac{1}{7.5} = \frac{1}{5} - \frac{2}{15}$.

Common denominator (15): $\frac{1}{v} = \frac{3}{15} - \frac{2}{15} = \frac{1}{15}$.

$v = +15$ cm.

Image position: The image is formed at 15 cm behind the mirror (since $v$ is positive). The object is between the pole (0 cm) and the focus (-7.5 cm). Expected image is virtual, erect, and magnified, behind the mirror.

Magnification $m = -v/u = -(+15 \text{ cm}) / (-5 \text{ cm}) = -15 / (-5) = +3$.

Nature of image: Since $v$ is positive, the image is formed behind the mirror, so it is a **virtual image**. Since $m$ is positive, the image is **erect**. Since $|m| = 3 > 1$, the image is **magnified**.

Position: 15 cm behind the mirror. Nature: Virtual, erect, magnified. Magnification: +3.

The side view mirror of a car is a convex mirror. Radius of curvature $R = 2$ m. For a convex mirror, $R$ is positive, so $R = +2$ m. Focal length $f = R/2 = +2/2 = +1$ m. The jogger is approaching, so object distance $u$ is decreasing. Let the jogger's initial distance be $u_1$. After time $\Delta t$, the distance is $u_2 = u_1 - v_{jogger} \Delta t$. The speed of the image is $|v_{image}| = |\Delta v / \Delta t|$, where $\Delta v$ is the change in image position. We can approximate the instantaneous speed of the image by differentiating the mirror equation with respect to time: $\frac{d}{dt}(\frac{1}{v} + \frac{1}{u} = \frac{1}{f})$.

$\frac{d}{dt}(v^{-1}) + \frac{d}{dt}(u^{-1}) = \frac{d}{dt}(f^{-1})$.

$-1 v^{-2} \frac{dv}{dt} - 1 u^{-2} \frac{du}{dt} = 0$ (since $f$ is constant).

$\frac{1}{v^2} \frac{dv}{dt} = -\frac{1}{u^2} \frac{du}{dt}$.

Here, $\frac{dv}{dt}$ is the speed of the image, and $\frac{du}{dt}$ is the speed of the object. The jogger's speed is $v_{jogger} = |\frac{du}{dt}| = 5$ m/s. The object is approaching, so $u$ is becoming less negative, so $\frac{du}{dt} = +5$ m/s (change in distance from pole per unit time in direction of incident light). So, $\frac{du}{dt}$ is positive if the object is moving towards the mirror from far away (conventionally object is placed to the left, incident light from left). Let's assume object is placed on the left ($u$ is negative) and moves towards the mirror, so its position coordinate becomes less negative with time, meaning $du/dt$ is positive. Thus $du/dt = 5$ m/s.

Speed of the image $\frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt}$. The term $(v/u)^2$ is the square of the lateral magnification $m^2$. $\frac{v}{u} = \frac{f}{u-f}$. So $\frac{dv}{dt} = -\left(\frac{f}{u-f}\right)^2 \frac{du}{dt}$.

Let's use $u$ as the object position coordinate. Incident light from left is positive. Object is in front, so $u$ coordinate is negative. Jogger moving towards mirror means $|u|$ is decreasing. $du/dt$ is positive if object moves towards positive x direction. Let mirror pole be at origin, principal axis along x-axis. Let incident light be from left to right (+x direction). Object is on the left, $u < 0$. Jogger moves towards mirror, so $u$ coordinate becomes less negative. $du/dt > 0$. $du/dt = +5$ m/s.

Convex mirror, $f = +1$ m. $u$ is negative. $v$ is positive (virtual image behind mirror). $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1} - \frac{1}{u} = \frac{u-1}{u}$. $v = \frac{u}{u-1}$. $\frac{dv}{dt} = \frac{d}{dt}\left(\frac{u}{u-1}\right) = \frac{(u-1)\frac{du}{dt} - u\frac{d(u-1)}{dt}}{(u-1)^2} = \frac{(u-1)\frac{du}{dt} - u\frac{du}{dt}}{(u-1)^2} = \frac{(u-1-u)\frac{du}{dt}}{(u-1)^2} = \frac{-1}{(u-1)^2}\frac{du}{dt}$.

Speed of the image (magnitude) $|v_{image}| = |\frac{dv}{dt}| = \frac{1}{(u-1)^2} |\frac{du}{dt}| = \frac{1}{(u-1)^2} \times 5$. Here $u$ is the negative object distance. Let the distance of the jogger from the mirror be $d = |u|$. So $u = -d$. Then $|v_{image}| = \frac{1}{(-d-1)^2} \times 5 = \frac{1}{(d+1)^2} \times 5$. The image is virtual ($v>0$), located between P and F ($0 < v < 1$ for $d>0$). As $d$ decreases (jogger approaches), $v = -d/(-d-1) = d/(d+1)$ increases (image moves towards F). $dv/dt$ is negative in the frame where image moves to the left. In car frame, image moves away from mirror. Speed is $|dv/dt|$.

Let's use the magnitude of distance $d$ in front of the mirror, so $u=-d$. $\frac{1}{v} + \frac{1}{-d} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} + \frac{1}{d} = \frac{d+f}{fd} \implies v = \frac{fd}{d+f}$. Here $f=+1$. $v = \frac{d}{d+1}$. As $d$ decreases, $v$ increases. $v_{image} = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{d(t)}{d(t)+1}\right) = \frac{\frac{dd}{dt}(d+1) - d\frac{dd}{dt}}{(d+1)^2} = \frac{\frac{dd}{dt}(d+1-d)}{(d+1)^2} = \frac{1}{(d+1)^2}\frac{dd}{dt}$.

Jogger approaches at 5 m/s, so $d$ decreases at 5 m/s. $\frac{dd}{dt} = -5$ m/s. Speed of image is $|\frac{dv}{dt}| = |\frac{1}{(d+1)^2}(-5)| = \frac{5}{(d+1)^2}$. $d$ is the distance of the jogger from the mirror.

Given $f=1$ m.

(a) Jogger is 39 m away, $d = 39$ m. Speed of image $= \frac{5}{(39+1)^2} = \frac{5}{40^2} = \frac{5}{1600} = \frac{1}{320} = 0.003125$ m/s.

(b) Jogger is 29 m away, $d = 29$ m. Speed of image $= \frac{5}{(29+1)^2} = \frac{5}{30^2} = \frac{5}{900} = \frac{1}{180} = 0.00555...$ m/s.

(c) Jogger is 19 m away, $d = 19$ m. Speed of image $= \frac{5}{(19+1)^2} = \frac{5}{20^2} = \frac{5}{400} = \frac{1}{80} = 0.0125$ m/s.

(d) Jogger is 9 m away, $d = 9$ m. Speed of image $= \frac{5}{(9+1)^2} = \frac{5}{10^2} = \frac{5}{100} = \frac{1}{20} = 0.05$ m/s.

The text's values $1/280, 1/150, 1/60, 1/10$ m/s are different. Let's re-read the problem carefully. "how fast the image of the jogger appear to move when the jogger is... away". This usually means the instantaneous speed at that specific distance. My derivative approach calculates this instantaneous speed. Let's re-check the text's sample calculation for u=-39. "after 1 s the position of the image v (for u = –39 + 5 = –34) is (34/35 )m". This seems to assume u becomes -34 after 1 sec, so speed is 5. v for u=-39 is 39/40 m. v for u=-34 is 34/35 m. Shift in position in 1s is $|(34/35) - (39/40)| = |(1360-1365)/1400| = |-5/1400| = 5/1400 = 1/280$ m. Average speed is shift/time = $(1/280)/1 = 1/280$ m/s. The text is calculating the *average* speed of the image over a 1-second interval starting from the stated distance. This is a valid way to interpret "how fast it appears to move" over a small time, but the instantaneous speed from the derivative is more precise.

Using the text's method (average speed over 1s starting from the given distance $d$): Object moves from $u_1=-d$ to $u_2=-(d-5)$. Image moves from $v_1 = \frac{-d}{-d+1} = \frac{d}{d-1}$ (using $u$ negative, $f$ positive for convex mirror, $1/v + 1/u = 1/f$. $1/v = 1/f - 1/u$. $u$ is distance magnitude, so $u=-d$. $1/v = 1/1 - 1/(-d) = 1+1/d = (d+1)/d$. $v=d/(d+1)$). Let's use the distances $d$ as in the original problem. Object distance $u = -d$. $v = \frac{-df}{(-d)-f} = \frac{-d(1)}{-d-1} = \frac{-d}{-d-1} = \frac{d}{d+1}$. $f=+1$. $v = \frac{d}{d+1}$. Okay, my $v$ formula was $d/(d+1)$ and earlier $d/(d+f)$ which simplifies to $d/(d+1)$ for $f=1$. Let's re-calculate $v$. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1} - \frac{1}{-d} = 1 + \frac{1}{d} = \frac{d+1}{d}$. So $v = \frac{d}{d+1}$. This is image position *behind* the mirror.

Let's use $u$ as the distance value from the mirror, $u>0$ in front. Then $u$ decreases with time, $du/dt = -5$. Mirror equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for convex mirror using standard formula for real/virtual image. But with Cartesian signs, $\frac{1}{v} + \frac{1}{-u} = \frac{1}{f}$. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$. $u$ is the distance in front, so $u>0$. But using Cartesian signs, $u$ is position coordinate, typically negative for object in front. Let's stick to Cartesian signs: $u_{coord} < 0$. $|u_{coord}|$ is distance $d$. So $u_{coord} = -d$. $d$ is decreasing. $du_{coord}/dt = +5$ m/s. $f=+1$. $\frac{1}{v} + \frac{1}{u_{coord}} = \frac{1}{f}$. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u_{coord}}$. $\frac{dv}{dt} = -\frac{v^2}{u_{coord}^2}\frac{du_{coord}}{dt}$. $|v_{image}| = |\frac{dv}{dt}| = \frac{v^2}{u_{coord}^2}|\frac{du_{coord}}{dt}| = (\frac{v}{u_{coord}})^2 \times 5$. $v = \frac{1}{1/f - 1/u_{coord}}$. Let's use the formula $|v_{image}| = \frac{f^2}{(u_{coord}-f)^2} |\frac{du_{coord}}{dt}| = \frac{1^2}{(u_{coord}-1)^2} \times 5$. $u_{coord} = -d$. $|v_{image}| = \frac{5}{(-d-1)^2} = \frac{5}{(d+1)^2}$. This is my instantaneous speed result.

The text's values suggest average speed over 1s. Using $d$ as the distance from the mirror.

(a) $d=39$. $u_1=-39$. $u_2=-(39-5)=-34$. $v_1 = \frac{1}{1 - 1/(-39)} = \frac{1}{1+1/39} = \frac{39}{40}$. $v_2 = \frac{1}{1 - 1/(-34)} = \frac{1}{1+1/34} = \frac{34}{35}$. Shift $= |v_2 - v_1| = |\frac{34}{35} - \frac{39}{40}| = |\frac{1360 - 1365}{1400}| = |-5/1400| = 5/1400 = 1/280$. Average speed $= 1/280$ m/s.

(b) $d=29$. $u_1=-29$. $u_2=-(29-5)=-24$. $v_1 = \frac{1}{1 - 1/(-29)} = \frac{29}{30}$. $v_2 = \frac{1}{1 - 1/(-24)} = \frac{24}{25}$. Shift $= |\frac{24}{25} - \frac{29}{30}| = |\frac{144 - 145}{150}| = |-1/150| = 1/150$. Average speed $= 1/150$ m/s.

(c) $d=19$. $u_1=-19$. $u_2=-(19-5)=-14$. $v_1 = \frac{1}{1 - 1/(-19)} = \frac{19}{20}$. $v_2 = \frac{1}{1 - 1/(-14)} = \frac{14}{15}$. Shift $= |\frac{14}{15} - \frac{19}{20}| = |\frac{56 - 57}{60}| = |-1/60| = 1/60$. Average speed $= 1/60$ m/s.

(d) $d=9$. $u_1=-9$. $u_2=-(9-5)=-4$. $v_1 = \frac{1}{1 - 1/(-9)} = \frac{9}{10}$. $v_2 = \frac{1}{1 - 1/(-4)} = \frac{4}{5}$. Shift $= |\frac{4}{5} - \frac{9}{10}| = |\frac{8 - 9}{10}| = |-1/10| = 1/10$. Average speed $= 1/10$ m/s.

The text's average speed calculation over 1s starting at the given distance matches the provided answers. So the question intends the average speed over a 1-second interval.

How fast the image of the jogger appear to move (average speed over 1s starting from the given distance): (a) 1/280 m/s, (b) 1/150 m/s, (c) 1/60 m/s, (d) 1/10 m/s.

The Earth rotates $360^\circ$ in 24 hours. When viewed from Earth, the Sun appears to complete a $360^\circ$ shift across the sky in approximately 24 hours (due to Earth's rotation). Assuming the Sun's apparent motion is uniform:

Time taken for $360^\circ$ shift = 24 hours.

To find the time taken for a $1^\circ$ shift, divide the total time by the total angle:

Time per degree = $\frac{24 \text{ hours}}{360^\circ}$.

Convert hours to minutes: 24 hours = $24 \times 60$ minutes = 1440 minutes.

Time per degree = $\frac{1440 \text{ minutes}}{360^\circ} = \frac{144}{36} \text{ minutes} = 4 \text{ minutes}$.

The Sun takes 4 minutes to shift by 1° when viewed from the Earth.

A real image is formed when light rays from an object point *actually converge* to a point after reflection or refraction. A virtual image is formed where the rays *appear to diverge* from a point.

For plane mirrors, the image is always virtual, erect, and of the same size as the object. The rays diverge from the virtual image point behind the mirror.

For convex mirrors, the image of a real object is always virtual, erect, diminished, and located between the pole and the focus behind the mirror. The rays diverge from the virtual image point behind the mirror.

However, the mirror equation $1/v + 1/u = 1/f$ is valid for both real and virtual objects and images, using the sign convention. Let's consider if a real image ($v$ positive) can be formed by these mirrors.

• Plane mirror: $f = \infty$. The mirror equation becomes $1/v + 1/u = 0 \implies v = -u$. For a real object in front of the mirror ($u$ negative), $v$ is positive, meaning the image is behind the mirror. Image is virtual. For a virtual object (rays converging towards a point behind the mirror, $u$ positive), $v$ is negative, meaning the image is in front of the mirror. In this case, a real image is formed. So, a plane mirror can form a real image of a virtual object.
• Convex mirror: $f > 0$. The mirror equation is $1/v + 1/u = 1/f$. For a real object in front ($u$ negative), $1/v = 1/f - 1/u = 1/f + 1/|u|$. Since $f>0$ and $|u|>0$, $1/v > 0$, so $v>0$. Image is behind the mirror, virtual. For a virtual object (rays converging towards a point behind the mirror, $u$ positive), $1/v = 1/f - 1/u$. If $u > f$, then $1/v = \frac{u-f}{uf} > 0$, $v>0$, image is virtual. If $u < f$, then $1/v = \frac{f-u}{uf} > 0$, $v>0$, image is virtual. If $u > f$, then $1/v = \frac{u-f}{uf}$. If $u > f$, $u-f > 0$, $v$ positive, image virtual. Is $u$ always positive for virtual object? Yes, point towards mirror is positive. So, if rays are converging towards a point behind the mirror, $u$ is positive distance behind mirror. $u$ is positive coordinate. $1/v = 1/f - 1/u$. Since $f>0$, if $u$ is large positive, $1/v$ is positive, $v$ positive (virtual). If $u$ is small positive (point very close behind mirror), $1/v$ could be negative. Let $u > 0$. If $u > f$, $1/v = (u-f)/uf$. If $u 0$, $v>0$ virtual. If $u=f$, $1/v=0$, $v=\infty$. If $u>f$, $1/v = (u-f)/uf$. If $u 0$, $v$ is positive (virtual). If $1/f - 1/u < 0$, $v$ is negative (real). This happens if $1/u > 1/f$, i.e., $u < f$.

Answer: Yes, both plane and convex mirrors can produce real images if the object is a **virtual object** (i.e., the incident rays are converging towards a point behind the mirror). For a plane mirror, a virtual object at a distance $|u|$ behind the mirror forms a real image at the same distance in front. For a convex mirror, a virtual object located closer to the mirror than its focal point (and behind the mirror, so $u>0$ and $u < f$) will form a real image in front of the mirror ($v<0$).

No, there is no contradiction. The statement that a virtual image cannot be caught on a screen means that the light rays forming the virtual image do not *actually converge* at the location of the virtual image. They only *appear to diverge* from that point. If you place a screen at the location of a virtual image, the light rays will pass through the screen or be blocked by it; they will not converge to form an image on the screen.

When we see a virtual image, for example, in a plane mirror or through a magnifying glass, our eye's lens acts as a converging lens. The diverging rays that appear to come from the virtual image enter our eye and are converged by the eye's lens onto the retina, which acts as a screen. Our brain then interprets the signals from the retina to perceive the image at the location of the virtual image.

So, the virtual image itself is not formed on a screen *at the location of the virtual image*. Instead, the eye's lens forms a *real image* of the virtual image *on the retina*. Our visual system's ability to converge diverging rays allows us to see virtual images. A typical screen placed at the virtual image location cannot perform this convergence.

Light rays travel from the fisherman (in air, rarer medium) to the diver (under water, denser medium). When light travels from a rarer medium to a denser medium, it bends **towards the normal**. Consider rays coming from the fisherman's head and feet. Let the fisherman be at a distance from the point directly above the diver. Rays from the top of the fisherman's head reach the water surface. They bend towards the normal upon entering the water and travel towards the diver. When the diver traces these refracted rays back into the air, they will appear to originate from a point higher than the fisherman's actual height.

Imagine the fisherman's head at point H above the water. Rays from H refract into the water towards the diver. Due to bending towards the normal, the rays entering the water will seem to come from a point H' higher than H when traced back from underwater. Similarly, rays from the fisherman's feet at F will refract and appear to come from a point F' lower than F (or higher than actual feet position relative to water level, but the apparent effect is related to the height difference). Because the rays from the top bend towards the normal and appear to originate from a higher point, the apparent height of the fisherman observed by the diver will be **taller** than the actual height.

This is the opposite effect of looking at an object underwater from air, where the object appears shallower than its real depth.

Yes, the apparent depth of a tank of water changes if viewed obliquely. The formula that apparent depth = real depth / refractive index ($h_{app} = h_{real} / n$) is strictly valid only for viewing near the normal (at small angles of incidence). When viewed obliquely (at larger angles of incidence), the apparent depth **decreases**. The bottom appears to be raised even more when viewed from an oblique angle compared to viewing near the normal.

Explanation: Consider a point at the bottom. Light rays from this point travel upwards. When viewed near the normal, the rays refract away from the normal upon entering the air, and tracing them back gives the apparent position closer to the surface. When viewed obliquely, the angle of incidence in water is larger. The angle of refraction in air is even larger. Tracing these highly deviated refracted rays back into the water, they intersect at a point *higher* than the apparent position seen near the normal. Thus, the apparent depth decreases when viewed obliquely.

Yes, this fact is of great use to a diamond cutter. The brilliance of a diamond is primarily due to **total internal reflection (TIR)**. TIR occurs when light travels from a denser medium to a rarer medium and the angle of incidence in the denser medium exceeds the critical angle ($i_c$). The critical angle is given by $\sin i_c = n_{rarer} / n_{denser}$.

Diamond has a very high refractive index ($n_{diamond} \approx 2.42$) compared to glass ($n_{glass} \approx 1.5$). For the diamond-air interface, the critical angle is $\sin i_c = 1/2.42 \approx 0.413$, which gives $i_c \approx 24.4^\circ$. This critical angle is very small compared to glass-air interface ($i_c = \sin^{-1}(1/1.5) \approx 41.8^\circ$).

A diamond cutter cuts the facets of the diamond in such a way that light entering the diamond undergoes multiple total internal reflections inside before emerging. Because the critical angle is so small, it is much easier for light to undergo TIR inside diamond than inside glass. By carefully cutting the facets at appropriate angles, the cutter can ensure that most of the light entering the diamond is reflected internally multiple times before exiting, making it sparkle brilliantly. The high refractive index also contributes to a large dispersion of light, further enhancing the display of colours (fire) when white light passes through a diamond. While high dispersion is also important, the small critical angle due to the high refractive index is key to maximising TIR.

Given: Point source in air ($n_1 = 1$). Spherical glass surface ($n_2 = 1.5$). Radius of curvature $R = 20$ cm. Distance of light source from glass surface $u = 100$ cm. Since the object is in front of the surface (assuming incident light direction is positive), $u = -100$ cm according to the Cartesian sign convention. The glass surface is convex if $R > 0$, concave if $R < 0$. From the example diagram, it appears to be a convex surface towards the incident light, so let's assume $R = +20$ cm.

We use the refraction formula for a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.

Substitute the given values: $n_1 = 1$, $n_2 = 1.5$, $u = -100$ cm, $R = +20$ cm.

$\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{+20}$.

$\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} = \frac{1}{40}$.

$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100} = \frac{5}{200} - \frac{2}{200} = \frac{3}{200}$.

$v = 1.5 \times \frac{200}{3} = \frac{3}{2} \times \frac{200}{3} = 100$ cm.

The image distance $v = +100$ cm. Since $v$ is positive, the image is formed in the same direction as the incident light (behind the spherical surface). The image is real.

The image is formed at a distance of 100 cm from the glass surface, behind the surface.

(i) Power of the lens: Given focal length $f = 0.5$ m. The lens is a glass lens, typically a convex lens for positive focal length. Power $P = 1/f$.

$P = \frac{1}{0.5 \text{ m}} = 2 \text{ m}^{-1} = 2 \text{ D}$. Since $f>0$, power is positive. The power of the lens is +2.0 dioptres.

(ii) Refractive index of glass: Double convex lens. Let light be incident from the left. The first surface is convex, so its center of curvature is to the right, $R_1 = +10$ cm $= +0.10$ m. The second surface is also convex, but its center of curvature is to the left (on the side from which light comes *from* the first surface), so $R_2 = -15$ cm $= -0.15$ m. Focal length $f = 12$ cm $= +0.12$ m. Let $n_1$ be the refractive index of the surrounding medium (air, $n_1=1$), and $n_2$ be the refractive index of glass.

Use the lens maker's formula: $\frac{1}{f} = (\frac{n_2}{n_1} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

$\frac{1}{0.12} = (\frac{n_2}{1} - 1) \left(\frac{1}{+0.10} - \frac{1}{-0.15}\right)$.

$\frac{1}{0.12} = (n_2 - 1) \left(\frac{1}{0.10} + \frac{1}{0.15}\right)$.

$\frac{1}{0.12} = (n_2 - 1) \left(\frac{100}{10} + \frac{100}{15}\right) = (n_2 - 1) \left(10 + \frac{20}{3}\right) = (n_2 - 1) \left(\frac{30 + 20}{3}\right) = (n_2 - 1) \left(\frac{50}{3}\right)$.

$\frac{1}{0.12} = \frac{50}{3} (n_2 - 1)$.

$n_2 - 1 = \frac{3}{50 \times 0.12} = \frac{3}{6.0} = 0.5$.

$n_2 = 1 + 0.5 = 1.5$.

The refractive index of glass is 1.5.

(iii) Focal length in water: Convex lens in air, $f_{air} = 20$ cm $= 0.20$ m. Refractive index of air $n_{air} = 1$, glass $n_{glass} = 1.5$. From lens maker's formula: $\frac{1}{f_{air}} = (n_{glass}/n_{air} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.

$\frac{1}{0.20} = (1.5/1 - 1) (\frac{1}{R_1} - \frac{1}{R_2}) = (0.5) (\frac{1}{R_1} - \frac{1}{R_2})$.

So, $(\frac{1}{R_1} - \frac{1}{R_2}) = \frac{1}{0.20 \times 0.5} = \frac{1}{0.10} = 10$ m$^{-1}$. This difference of curvatures is 10 m$^{-1}$.

Now, the same lens is placed in water. Refractive index of water $n_{water} = 1.33$. Refractive index of glass relative to water is $n_{glass}/n_{water} = 1.5/1.33 \approx 1.128$. Let $f_{water}$ be the focal length in water.

$\frac{1}{f_{water}} = (n_{glass}/n_{water} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.

$\frac{1}{f_{water}} = (1.128 - 1) \times (10 \text{ m}^{-1}) = (0.128) \times 10 = 1.28$ m$^{-1}$.

$f_{water} = \frac{1}{1.28} \text{ m} \approx 0.781 \text{ m} = 78.1$ cm. The text gives 78.2 cm.

The focal length of the convex lens in water is approximately 78.1 cm.

We have three thin lenses along an axis. Lens 1: Convex, $f_1 = +10$ cm. Lens 2: Concave, $f_2 = -10$ cm. Lens 3: Convex, $f_3 = +30$ cm. The object is placed at 30 cm in front of the first lens. Let's denote distances and positions relative to each lens.

Step 1: Image formed by the first lens (L1). Object distance $u_1 = -30$ cm (in front of L1). $f_1 = +10$ cm.

Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$.

$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{+10}$.

$\frac{1}{v_1} + \frac{1}{30} = \frac{1}{10}$.

$\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$.

$v_1 = +15$ cm. The image $I_1$ is formed 15 cm to the right of the first lens. This is a real image.

Step 2: Image formed by the second lens (L2). The second lens is located 5 cm to the right of the first lens. The image $I_1$ formed by L1 is at 15 cm to the right of L1. So, the distance of $I_1$ from L2 is $u_2 = 15 \text{ cm} - 5 \text{ cm} = 10 \text{ cm}$. Since $I_1$ is on the right side of L2, and incident light on L2 is from the left, $I_1$ acts as a virtual object for L2. By sign convention, object distance for a virtual object is positive. So $u_2 = +10$ cm. $f_2 = -10$ cm.

Using the lens formula for L2: $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$.

$\frac{1}{v_2} - \frac{1}{+10} = \frac{1}{-10}$.

$\frac{1}{v_2} = \frac{1}{10} - \frac{1}{10} = 0$.

$v_2 = \infty$. The image $I_2$ is formed at infinity.

Step 3: Image formed by the third lens (L3). The third lens is located 10 cm to the right of the second lens. The image $I_2$ formed by L2 is at infinity. This serves as the object for L3. The object is at infinity, so $u_3 = \infty$. $f_3 = +30$ cm.

Using the lens formula for L3: $\frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3}$.

$\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{+30}$.

$\frac{1}{v_3} - 0 = \frac{1}{30}$.

$v_3 = +30$ cm.

The final image $I_3$ is formed 30 cm to the right of the third lens. Since $v_3$ is positive, the final image is real.

The position of the final image is 30 cm to the right of the third lens.

The text has already worked through this example and found the final image position is 30 cm to the right of the third lens. Please refer back to Example 9.9 in the main text for the solution steps.

This refers to Example 9.9, which was already worked through. The final image is formed 30 cm to the right of the third lens.

This refers to Example 9.9 which was worked through already. The final image is formed 30 cm to the right of the third lens.