Ray Optics and Optical Instruments: common doubts

Question 1

Why is the focal length of a concave mirror taken as negative?

Accepted Answer

Because of the Cartesian sign convention: distances are measured from the pole, positive in the direction the incident light travels and negative against it. With light coming from the left, a concave mirror's focus lies in front of it, on the side the light comes from, so its f and R are negative. A convex mirror's focus lies behind it, so its f and R are positive.

Question 2

Why is the focal length of a spherical mirror half its radius of curvature?

Accepted Answer

For rays close to the principal axis, simple geometry gives f = R/2. A ray parallel to the axis reflects so that the radius at the point of incidence, which is the normal, makes equal angles with both rays. For small angles this sends the reflected ray through the point midway between the pole and the centre of curvature. Rays far from the axis miss that point slightly, so the result holds only for paraxial rays.

Question 3

How do I avoid sign mistakes when using the mirror formula?

Accepted Answer

Keep the formula in its standard form, 1/v + 1/u = 1/f, and put signs only into the numbers you substitute. A real object in front of a mirror has negative u, and a concave mirror of focal length 20 cm has f = −20 cm. A negative answer for v then means a real image in front of the mirror, a positive one a virtual image behind it. Adding signs to the formula itself counts them twice.

Question 4

What does a negative magnification mean for a mirror?

Accepted Answer

A negative magnification means the image is inverted, and for a real object it is also real. With m = −v/u, a real image in front of the mirror has u and v both negative, which makes m negative. A positive m means an erect image, which for a real object is virtual. The size of m, ignoring its sign, tells whether the image is enlarged, if more than 1, or diminished, if less.

Question 5

Can a convex mirror ever form a real image?

Accepted Answer

Not of a real object. With f positive and u negative, the mirror equation always gives a positive v, so the image is virtual, erect, smaller than the object and behind the mirror, wherever the object is. A convex mirror can give a real image only of a virtual object, light already converging towards a point behind the mirror, and only when that point lies between the pole and the focus.

Question 6

Why does a swimming pool look shallower than it really is?

Accepted Answer

Because light from the bottom bends away from the normal as it leaves the water, so it seems to come from a point higher up. Seen from nearly straight above, the apparent depth is the real depth divided by the refractive index. For water, with n about 1.33, a pool 2 m deep looks only about 1.5 m deep. Looking at a slant makes the effect even stronger.

Question 7

How do I find the apparent depth of a coin under two layers of different liquids?

Accepted Answer

Treat each layer separately and add the results: seen from nearly straight above, apparent depth = t₁/n₁ + t₂/n₂ + …, where t is a layer's thickness and n its refractive index. Each layer shortens only its own thickness, by its own index. Adding the thicknesses first and dividing by one index gives a wrong answer unless the two liquids have the same refractive index.

Question 8

What is the difference between relative and absolute refractive index?

Accepted Answer

Absolute refractive index compares the speed of light in vacuum with its speed in the medium, n = c/v. Relative refractive index compares two media: n₂₁ = n₂/n₁ = v₁/v₂, the index of medium 2 with respect to medium 1. Snell's law, sin i/sin r = n₂₁, uses the relative value, so for light passing from water into glass you need n_glass/n_water, not n_glass alone.

Question 9

What conditions are needed for total internal reflection?

Accepted Answer

Light must travel from a denser medium towards a rarer one, and its angle of incidence must be greater than the critical angle. Going from air into glass, light always bends towards the normal and some always enters, so total internal reflection cannot happen. The critical angle follows from sin i_c = 1/n, with n the denser medium's index relative to the rarer one; for glass in air it is about 42°.

Question 10

How can the critical angle be found from the speeds of light in two media?

Accepted Answer

Use sin i_c = v_denser/v_rarer. Since the refractive index is n = c/v, the ratio of the indices of the two media equals the inverse ratio of the speeds. The critical-angle relation sin i_c = 1/n can therefore be written with speeds instead. If light travels at 2 × 10⁸ m/s in glass and 3 × 10⁸ m/s in air, sin i_c = 2/3.

Question 11

How do optical fibres carry light along a bent path?

Accepted Answer

By total internal reflection at the boundary between the core and the surrounding cladding, which has a lower refractive index. Light entering the core meets this boundary at angles larger than the critical angle, so it is reflected back completely each time, again and again along the fibre. Very little light is lost, which is why fibres are used in telecommunication and to look inside the body.

Question 12

When should I use the formula for refraction at a single spherical surface?

Accepted Answer

Use n₂/v − n₁/u = (n₂ − n₁)/R when light crosses a single curved boundary between two media, as for an object inside a glass sphere or a fish in a round bowl. Here n₁ is the medium the light starts in and n₂ the one it enters, and R is positive when the centre of curvature lies on the side the refracted light goes. For a thin lens, the lens formula is quicker.

Question 13

Why does a convex lens become diverging when placed in a denser liquid?

Accepted Answer

Because the lens maker's formula uses the lens's refractive index relative to its surroundings, n₂₁ = n_lens/n_medium. If the liquid is optically denser than the glass, n₂₁ is less than 1, so (n₂₁ − 1) turns negative and the focal length changes sign. In a liquid of exactly the same index, n₂₁ = 1 and the lens has no focusing power at all.

Question 14

What is the difference between the mirror formula and the lens formula?

Accepted Answer

For a mirror, 1/v + 1/u = 1/f with magnification m = −v/u; for a thin lens, 1/v − 1/u = 1/f with m = v/u. The difference arises because a mirror sends light back, so its real images form on the object's side, while a lens lets light through, so its real images form on the far side. Using one formula for the other device is a frequent exam slip.

Question 15

What happens to the image if half of a lens is covered?

Accepted Answer

The full image still forms, in the same place and of the same size, but it is dimmer. Every part of the lens receives light from every point of the object and can form a complete image, so the uncovered half does the whole job with half the light. The focal length is unchanged, since it depends only on the curvature of the surfaces and the refractive index.

Question 16

Why do the powers of thin lenses in contact add, but not their focal lengths?

Accepted Answer

Because the image formed by the first lens acts as the object for the second, and combining the two lens equations gives 1/f = 1/f₁ + 1/f₂. Power is P = 1/f, so the powers simply add: P = P₁ + P₂. Give a diverging lens a negative focal length and a negative power before adding. Power comes out in dioptres only when focal length is in metres.

Question 17

What is one dioptre?

Accepted Answer

One dioptre is the power of a lens whose focal length is one metre: 1 D = 1 m⁻¹. Since P = 1/f with f in metres, a converging lens of focal length 25 cm has power +4 D, and a diverging lens of focal length 50 cm has power −2 D. A larger power means the lens bends light more strongly and has a shorter focal length.

Question 18

Why does the deviation produced by a prism first decrease and then increase?

Accepted Answer

Because there is one angle of incidence at which the ray passes through the prism symmetrically, and there the deviation is least. With δ = i + e − A, the deviation falls as i grows from small values, reaches its minimum D_m when i = e and r₁ = r₂ = A/2, then rises again. At minimum deviation, n₂₁ = sin[(A + D_m)/2] / sin(A/2) gives the refractive index.

Question 19

When can I use D_m = (n − 1)A for a prism?

Accepted Answer

Only for a thin prism, one whose refracting angle is a few degrees. For small angles the sines in the full formula can be replaced by the angles themselves, which reduces it to D_m = (n − 1)A. For a 60° prism this shortcut gives badly wrong answers, so use n = sin[(A + D_m)/2] / sin(A/2) instead.

Question 20

What is the difference between magnifying power with the image at the near point and at infinity?

Accepted Answer

With the final image at the near point, 25 cm from the eye, a simple magnifier gives m = 1 + D/f; with the image at infinity, so the eye is relaxed, it gives m = D/f. The near-point setting gives slightly more magnification but tires the eye. The same choice applies to the eyepiece of a compound microscope, so always check which adjustment a question means.

Question 21

Why does a compound microscope use an objective and an eyepiece of short focal length?

Accepted Answer

Because both magnifications rise as the focal lengths fall. The objective forms a real, enlarged image with m_o = L/f_o, and the eyepiece magnifies that image again; with the final image at infinity, m = (L/f_o)(D/f_e). Here L, the tube length, runs from the objective's second focal point to the eyepiece's first focal point, not between the lenses themselves.

Question 22

Why does an astronomical telescope need a long-focus objective and a short-focus eyepiece?

Accepted Answer

Because its magnifying power in normal adjustment is m = f_o/f_e, so a long objective focal length and a short eyepiece focal length both raise it. The tube is then f_o + f_e long and the final image is inverted. Large telescopes often use a concave mirror as the objective instead, which avoids chromatic aberration and is lighter and easier to support than a big lens.

Question 23

Does the focal length of a mirror change when it is placed in water, as a lens's does?

Accepted Answer

No. A mirror's focal length, f = R/2, depends only on its curvature, because the law of reflection is the same in every medium. A lens's focal length depends on the refractive index of the glass relative to its surroundings, so a glass lens placed in water has a longer focal length, and a smaller power, than in air. So a mirror and a lens of equal focal length in air behave differently in water.

Ray Optics and Optical Instruments: common doubts, answered

About the chapter

Does the focal length of a mirror change when it is placed in water, as a lens's does?

Reflection by spherical mirrors and the sign convention