Time taken for 360° shift = 24 h
Time taken for 1° shift = 24/360 h = 4 min.

The ray diagram for the formation of the image of the phone is shown in Fig. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B'C = BC. You can yourself realise why the image is distorted.

You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).

Size of the object, h₁ = 3 cm
Object distance, u = -14 cm
Focal length of the concave lens, f = -21 cm
Image distance = v
According to the lens formula, we have the relation:
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =-\frac { 1 }{ 21 } -\frac { 1 }{ 14 } =\frac { -2-3 }{ 42 } =\frac { -5 }{ 42 } \)
\(\therefore v=-\frac { 42 }{ 5 } =-84cm\)
Hence, the image is formed on the other side of the lens, 8.4 cm away from it. The negative sign shows that the image is erect and virtual.
The magnification of the image is given as:
\(m=\frac { Image \ height({ h }_{ 2 }) }{ Object \ height({ h }_{ 1 }) } =\frac { -8.4 }{ -14 } \)
\(\therefore { h }_{ 2 }=\frac { -8.4 }{ -14 } \times 3=0.6\times 3=1.8\) cm
Hence, the height of the image is 1.8 cm
If the object is moved further away from the lens, then the virtual image will move toward the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.

Actual depth of the bulb in water, d₁ = 80 cm = 0.8 m
Refractive index of water, μ = 1.33
The given situation is shown in the following figure:

Where,
i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle of radius, \(R=\frac { AC }{ 2 } =OA=OB\)
Using Snell’ law, we can write the relation for the refractive index of water as:
\({ \mu }=\frac { sin \ i }{ sin \ r } \)
\(1.33=\frac { sin{ 90 }^{ o } }{ sini } \)
\(\therefore i={ sin }^{ -i }\left( \frac { 1 }{ 1.33 } \right) =48.{ 75 }^{ o }\)
Using the given figure, we have the relation:
\(tan \ i=\frac { OC }{ OB } =\frac { R }{ { d }_{ 1 } } \)
∴ R = tan 48.75° × 0.8 = 0.91 m
∴ Area of the surface of water = πR² = π (0.91)² = 2.61 m²
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m².

Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
\(f=\frac { R }{ 2 } =-18\)cm
Image distance = v
The image distance can be obtained using the mirror formula:
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)
\(\frac { 1 }{ -18 } =\frac { 1 }{ -27 } =\frac { -3+2 }{ 54 } =-\frac { 1 }{ 54 } \)
∴ v = -54 cm
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
\(m=\frac { { h }^{ ' } }{ h } =-\frac { v }{ u } \)
\(\therefore { h }^{ ' }=-\frac { v }{ u } \times h\)
\(=-\left( \frac { -54 }{ -27 } \right) \times 2.5=-5\)cm
The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.

From the mirror equation, Eq., we get \(v=\frac{f u}{u-f}\)
For convex mirror, since \(R=2 \mathrm{~m}, f=1 \mathrm{~m}\). Then for \(u=-39 \mathrm{~m}, v=\frac{(-39) \times 1}{-39-1}=\frac{39}{40} \mathrm{~m}\)
Since the jogger moves at a constant speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\), after 1 s the position of the image v (for \(u=-39+5=-34)\) is (34 / 35) m.
The shift in the position of image in 1 s is \(\frac{39}{40}-\frac{34}{35}=\frac{1365-1360}{1400}=\frac{5}{1400}=\frac{1}{280} \mathrm{~m}\)
Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) m s^–1 Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is
\(\frac{1}{150} \mathrm{~m} \mathrm{~s}^{-1}, \frac{1}{60} \mathrm{~ms}^{-1} \text { and } \frac{1}{10} \mathrm{~ms}^{-1} \text {, respectively. }\)
Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.

The focal length f = -15/2 cm = -7.5 cm
(i) The object distance u = -10 cm. Then Eq gives
\(\frac { 1 }{ v } +\frac { 1 }{ 10 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 10\times 7.5 }{ -2.5 } =-30\) cm
The image is 30 cm from the mirror on the same side as the object
Also, magnification m = \(\frac { v }{ u } =-\frac { (-30) }{ (-10) } =-3\)
The image is magnified, real and inverted.
(ii) The object distance u = -5 cm. Then from Eq
\(\frac { 1 }{ v } +\frac { 1 }{ -5 } =\frac { 1 }{ -7.5 } \)
or \(v=\frac { 5\times 7.5 }{ (7.5-5) } =15\) cm
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = 15 \(-\frac { v }{ u } =-\frac { 15 }{ (-5) } =3\)
The image is magnified, virtual and erect.

Refractive index of glass, μ
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R₁
Radius of curvature of the other face of the lens = R₂
Radius of curvature of the double-convex lens = R The value of R can be calculated as:
∴ R₁ = R and R₂ = -R
The value of R can be calculated as:
\(\frac { 1 }{ f } =(\mu -1)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
\(\frac { 1 }{ 20 } =(1.55)\left[ \frac { 1 }{ R } +\frac { 1 }{ R } \right] \)
\(\frac { 1 }{ 20 } =0.55\times \frac { 2 }{ R } \)
∴ R = 0.55 x 2 x 20 = 22 cm
Hence, the radius of curvature of the double-convex lens is 22 cm.