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9.18 (a) Yes, Plane and convex mirrors can produce real images as well, if the object is virtual.

i.e. if the light rays converging at a point behind a plane mirror or a convex mirror, they are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No. A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

(c) Water being the denser medium than air, the light rays will deviate and the fisherman will appear to be taller to the diver.

(d) The apparent depth of the tank water changes when viewed obliquely. This is because light bends on travelling from one medium to another. The apparent depth of the tank when viewed obliquely is less than the near normal viewing.

(e) No, diamond is used as cutter because of its hardness, not refractive index.

9.17 Refractive index of the glass fibre, ${\mu }_1$ = 1.68

Refractive index of outer covering of the pipe, ${\mu }_2$ = 1.44

Given:-

Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’

The refractive index of the inner core-outer core interface, $\mu$ is given as

$\mu =\frac{{\mu }_2}{{\mu }_1}$ = $\frac{1}{\sin ?i\text{'}}$

$\frac{1}{\sin ?i\text{'}}=\frac{1.44}{1.68}$ or $\sin ?i\text{'}$ = $\frac{1.44}{1.68}$

i’ = 59 $°$

For the critical angle, total reflection take place only when i > i’ i.e. i > 59 $°$

Maximum angle of reflection, ${\tau }_{max}$ = 90 $°-i\text{'}$ = 31 $°$

Let $i_{max}$ be the maximum angle of incidence and $r_{max}$ be the maximum angle of reflection.

${\mu }_1=\frac{\sin ?i_{max}}{\sin ?r_{max}}$ or $\sin ?i_{max}={\mu }_1\times \sin ?r_{max}$

$\sin ?i_{max}=$ 1.68 $\times sin31°$

$i_{max}=59.91°$ $\approx 60°$

The entire rays incident at angles lying in the range of 0 < i < 60 will suffer total internal reflection.

If the outer covering is absent, then:

Refractive index of the outer pipe, ${\mu }_1$ = Refractive index of air = 1

For the angle of incidence, i = 90 $°$ , we can write Snell’s law at ‘air – pipe’ interface as :

$\frac{sini}{\sin ?r}$ = ${\mu }_2$ = 1.68

$\sin ?r=$ $\frac{\sin ?i}{1.68}$ = $\frac{\sin ?90°}{1.68}$ = $\frac{1}{1.68}$

r = 36.53 $°$

i’ = 90 $°-$ 36.53 $°$ = 53.47 $°$

Since i’ > r, all incident rays will suffer total internal reflection.

9.15 For a concave mirror, the focal length, f < 0

When the object is placed on the left side of the mirror, the object distance, u is negative

For image distance v, we can write the lens formula as:

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$ (since u is negative) ……….(1)

The object lies between f and 2f, i.e. 2f < u < f

or, $\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$

or, $-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$

or, $\frac{1}{f}$ $-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}$ < 0 …………(2)

Using equation (1), we get

$\frac{1}{2f}<\frac{1}{v}$ < 0

Therefore, $\frac{1}{v}$ is negative, hence v is negative

$\frac{1}{2f}<\frac{1}{v}or$ 2f > v or –v > -2f . Hence, the image lies beyond 2f

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

Using equation (2), we can conclude that:

$\frac{1}{v}<0$ or v > 0

Thus, the image is formed on the back side of the mirror.

Hence, a convex mirror always produces a virtual image, regardless of the object distance.

For a convex mirror, the focal length, the focal length f > 0

When the object is placed on the left side of the mirror, the object distance (u) is negative (< 0)

For image distance v, we have the mirror formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

But we have u < 0, then

$\frac{1}{v}>\frac{1}{f}$ or v < f

$\mathrm{H}$ ence, the image formed is diminished and it is located between the focus (f) and the pole.

For a concave mirror, the focal length (f) is negative, f < 0

When the object is placed on the left side of the mirror, the object distance (u) is negative, u < 0

It is placed between the focus (f) and the pole, so f > u > 0

So, $\frac{1}{f}$ < $\frac{1}{u}$ < 0 or $\frac{1}{f}$ - $\frac{1}{u}$ < 0

For image distance v, we have the formula:

$\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}$ < 0 or v > 0

The image is formed on the right side of the mirror, hence it is a virtual image.

For u < 0 and v > 0, we can write $\frac{1}{u}$ > $\frac{1}{v}$ or v > u

Magnification, m = $\frac{v}{u}$ > 1

Hence, the formed image is enlarged.

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9.14 Focal length of the objective lens, $f_o$ = 15 m = 1500 cm

Focal length of the eyepiece, $f_e$ = 1.0 cm

The angular magnification of the telescope is given by,

m = $\frac{f_o}{f_e}$ = $\frac{1500}{1}$ = 1500

Hence the angular magnification of the telescope is 1500

Diameter of the moon, $d_m$ = 3.48 $*{10}^6$ m

Radius of the lunar orbit, $r_l$ = 3.8 $*{10}^8$ m

Let $d_1$ be the diameter of the image of the moon formed by the objective lens.

The angle subtended by the diameter of the moon is equal to the angle subtended by the image. Hence,

$\frac{d_m}{r_l}$ = $\frac{d_1\mathrm{}}{f_o\mathrm{}}$

$d_1=\frac{d_m*f_o}{\mathrm{}{r}_l}$ = $\frac{3.48\mathrm{}*{10}^6*1500\mathrm{}}{3.8\mathrm{}*{10}^8}$ = 13.74 cm

Hence the diameter of the moon's image is 13.74 cm.