10th Science Unit 2 Optics Book Back Answers

Refractive index $\mu = c/v$, so speed $v = c/\mu$.

The smaller the refractive index, the greater the speed of light in that medium.

Among 1.31, 1.43, 1.33 and 2.4, the least value is 1.31 which belongs to substance A.

For a convex lens, when the object is placed at the centre of curvature C (that is at 2f), a real and inverted image of the same size is formed at the other centre of curvature.

Rays starting from the principal focus of a convex lens emerge parallel to the principal axis (Rule-3 of ray diagrams).

So the bulb at the focus gives a parallel beam after refraction.

A convex lens forms real inverted images for most object positions, and a virtual erect image when the object lies between F and O.

Hence its magnification can be either positive (virtual, erect) or negative (real, inverted).

When an object is placed at infinity, a convex lens forms a real, point-sized (highly diminished) image at the principal focus.

Power of a lens $P = \dfrac{1}{f}$, so $f = \dfrac{1}{P}$.

$f = \dfrac{1}{-4} = -0.25$ m (negative sign shows it is a concave lens).

In myopia the eye ball lengthens or the focal length of the eye lens decreases.

So the image of a distant object is formed in front of (before) the retina; it is corrected using a concave lens.

In presbyopia the ciliary muscles become weak and the eye lens becomes rigid, so the eye loses its power of accommodation and both distant and near vision are affected.

Such combined defects are corrected by bifocal lenses — concave in the upper part for distant vision and convex in the lower part for reading.

Small letters are read using a convex lens acting as a simple microscope (magnifying lens).

Magnifying power increases when the focal length is smaller, so a convex lens of focal length 5 cm is preferred.

Blue light has the shorter wavelength and the highest refractive index in glass; red has the longest wavelength and the lowest refractive index.

Since $v = c/\mu$, higher refractive index means lower speed.

Therefore $V_B < V_G < V_R$.

The textbook defines: the path of light is called a ray of light, and a group of such rays is a beam of light.

Refractive index $\mu = c/v$, and the speed of light in any medium is less than in vacuum.

So the ratio is always greater than 1.

When the energy of the incident beam and the scattered beam are the same, no energy is lost in the interaction.

Such scattering is called elastic scattering (e.g. Rayleigh and Mie scattering).

Rayleigh's law: amount of scattering $S \propto \dfrac{1}{\lambda^4}$.

Hence shorter wavelength colours (blue) scatter more than longer wavelength colours (red).

The iris is the coloured part of the eye and it controls the amount of light entering through the pupil, like a camera aperture.

Refraction happens because light travels at different speeds in different media.

The textbook states that the velocity of light is more in a rarer medium and less in a denser medium.

So the statement is wrong; the speeds are the other way round.

Power of a lens is the degree of convergence or divergence of light rays.

It is numerically the reciprocal of the focal length, $P = \dfrac{1}{f}$.

Hence power directly depends on the focal length of the lens.

Hypermetropia occurs when the focal length of the eye lens is increased (or the eye ball is shortened), so the converging power of the eye lens decreases.

The image of nearby objects is then formed behind the retina.

An increase in converging power (shorter focal length) causes myopia, not hypermetropia.

A convex lens produces mostly real and inverted images for objects placed beyond the principal focus.

It gives a virtual image only when the object is placed between the principal focus F and the optical centre O, and that image is enlarged, not small.

So a convex lens does not always give a small virtual image.

Retina is the back surface of the eye on which the real and inverted image is formed — it acts as the screen of the eye: 1 - d

Pupil is the centre part of the iris and is the pathway for light to reach the retina: 2 - a

Ciliary muscles change the focal length of the eye lens to focus near and distant objects — this is the power of accommodation: 3 - f

In myopia the far point is no longer infinity; it has come closer: 4 - b

In hypermetropia the near point is no longer 25 cm; it has moved away (farther): 5 - c

Refractive index is defined as $\mu = \dfrac{c}{v}$, where $c$ is the speed of light in vacuum and $v$ is its speed in the medium.

From this relation, $\mu \propto \dfrac{1}{v}$ — so a medium with a high refractive index (denser medium) has a low speed of light in it. Hence the assertion is true.

The reason states exactly this inverse relation, which is true, and it is the very reason why light is slower in a denser medium.

So both assertion and reason are true, and the reason correctly explains the assertion.

In myopia the eye ball lengthens and the focal length of the eye lens is reduced, i.e. its converging power increases, so the image of a distant object is formed in front of the retina. Hence the assertion is true.

Myopia is corrected by using a concave (diverging) lens, which moves the image back on to the retina. Hence the reason is also true.

But the reason only tells how the defect is corrected; it does not explain why myopia is caused (the increased converging power).

So both statements are true, but the reason is not the correct explanation of the assertion.

Refractive index tells us how fast or how slow light travels in a medium.

It is defined as the ratio of the speed of light in vacuum (c) to the speed of light in that medium (v): $\mu = \dfrac{c}{v}$.

If the refractive index of a medium is high, the speed of light in that medium is low and vice versa.

It has no unit, because it is a ratio of two speeds.

Snell's law is the second law of refraction.

It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.

In symbols, $\dfrac{\sin i}{\sin r} = \dfrac{\mu_2}{\mu_1}$, i.e. $\mu_1 \sin i = \mu_2 \sin r$.

This ratio is a constant for a given pair of media and a given colour of light.

Sunlight is a composite light made of light of different colours (wavelengths).

When a beam of white or composite light is refracted through a transparent medium such as glass or water, it splits into its component colours.

This splitting of white light into its constituent colours is called dispersion of light; the band of colours obtained is called the spectrum (VIBGYOR).

It happens because the refractive index of the medium is different for different wavelengths, so each colour bends through a different angle — least for red and most for violet.

Rayleigh's law deals with the scattering of sunlight by the atoms and molecules of gases in the atmosphere.

It states: “The amount of scattering of light is inversely proportional to the fourth power of its wavelength.”

In symbols, $S \propto \dfrac{1}{\lambda^4}$.

Hence shorter wavelength colours (blue, violet) are scattered much more than longer wavelength colours (red).

Shape: a convex lens is thicker in the middle than at the edges; a concave lens is thinner in the middle than at the edges.

Action on light: convex lens converges a parallel beam (converging lens); concave lens diverges the beam (diverging lens).

Image: convex lens produces mostly real images; concave lens produces virtual images.

Use in eye defects: convex lens corrects hypermetropia; concave lens corrects myopia.

Power: power of a convex lens is positive; power of a concave lens is negative.

The ability of the eye lens to focus both nearby objects and distant objects clearly on the retina is called the power of accommodation.

It is achieved by changing the focal length of the eye lens with the help of the ciliary muscles.

For distant objects the ciliary muscles relax, the lens becomes thinner and its focal length increases.

For nearby objects the ciliary muscles contract, the lens becomes thicker and its focal length decreases.

Myopia (short sightedness) is caused by the lengthening of the eyeball, i.e. the distance between the eye lens and the retina increases.

It is also caused by the decrease in the focal length of the eye lens (excess converging power).

Because of this, the image of a distant object is formed in front of the retina instead of on it.

So the far point is no longer at infinity; it comes closer, and distant objects look blurred while nearby objects are seen clearly.

It is corrected by using a concave lens of suitable focal length.

Sunlight entering the atmosphere is scattered by the atoms and molecules of the gases present (Rayleigh scattering).

By Rayleigh's law, the amount of scattering $S \propto 1/\lambda^4$, so shorter wavelengths are scattered much more.

Blue light has a shorter wavelength than red light, so blue is scattered to a much greater extent.

This scattered blue light reaches our eyes from all directions of the sky, and hence the sky appears blue.

By Rayleigh's law, the amount of scattering is inversely proportional to the fourth power of the wavelength.

Red light has the highest wavelength among visible colours, so it is scattered the least.

Being least scattered, red light can travel the longest distance through air, fog, mist or smoke without being lost.

Hence a red signal is visible from a great distance to drivers, and red is used for traffic signals (danger/stop).

(a) Yes, he can still get the image. A lens broken along its principal axis still has the same two spherical refracting surfaces, so each half continues to refract the light rays and form the image at the same position.

However, the aperture of the broken lens is reduced, so fewer rays are collected. The image formed will be less bright (dimmer) than before, but its position, size and nature remain the same.

(b) No, there is no change in the focal length.

From the lens maker's formula $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, the focal length depends only on the refractive index $\mu$ of the material and the radii of curvature $R_1$ and $R_2$. Breaking the lens along the axis does not change $\mu$, $R_1$ or $R_2$, so $f$ stays the same.

Nocturnal birds like the owl hunt at night, when the intensity of light available is very low.

The cornea is the main refracting surface of the eye; a large cornea can receive and bend a larger amount of the faint light coming from the object.

The pupil is the pathway through which light reaches the retina. A large pupil allows much more light to enter the eye.

With more light falling on the retina, a bright and clear real, inverted image of the object is formed even in dim light, giving the owl excellent night vision.