The following figure shows a Cassegrain telescope consisting of a concave mirror and a convex mirror.

Distance between the objective mirror and the secondary mirror, d = 20 mm
Radius of curvature of the objective mirror, R₁ = 220 mm
Hence, focal length of the objective mirror,  \({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =110\)
Radius of curvature of the secondary mirror, R₁ = 140 mm
Hence, focal length of the secondary mirror, \({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } \) = 70 mm
The image of an object placed at infinity, formed by the objective mirror, will act as a virtual object for the secondary mirror.
Hence, the virtual object distance for the secondary mirror, u = f₁ - d
= 110 - 20
= 90 mm
Applying the mirror formula for the secondary mirror, we can calculate image distance (v) as:
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \)
\(\frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \)
\(\frac { 1 }{ 70 } -\frac { 1 }{ 90 } =\frac { 9-7 }{ 630 } =\frac { 2 }{ 630 } \)
∴ v = \(\frac{630}{2}\) = 315 mm
Hence, the final image will be formed 315 mm away from the secondary mirror.

Focal length of the objective lens, f_o = 1.25 cm
Focal length of the eyepiece, f_e = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
\({ m }_{ e }=\left( 1+\frac { d }{ { f }_{ e } } \right) \)
\(=\left( 1+\frac { 25 }{ 5 } \right) =6\)
The angular magnification of the objective lens (m_o) is related to m_e as:
m_o m_e = m
m_o = \(\frac { m }{ { m }_{ c } } \)
\(=\frac { 30 }{ 6 } \) = 5
We also have the relation:
m_o = \(\frac{Image \ distance \ for \ the \ objective \ lens (v_o)}{ Object \ distace \ for \ the \ objective \ lens (u_o)}\)
\(5=\frac { { v }_{ o } }{ -{ u }_{ o } } \)
∴ vo = -5u_o  .........(1)
Applying the lens formula for the objective lens:
\(\frac { 1 }{ { f }_{ o } } =\frac { 1 }{ { v }_{ o } } -\frac { 1 }{ { u }_{ o } } \)
\(\frac { 1 }{ 1.25 } =\frac { 1 }{ -5{ u }_{ o } } -\frac { 1 }{ { u }_{ o } } =\frac { -6 }{ 5{ u }_{ 0 } } \)
\(\therefore { u }_{ o }=\frac { -6 }{ 5 } \times 1.25=-1.5\) cm
And v_o = -5u_o
= -5 x (-1.5) = 7.5 cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
\(\frac { 1 }{ { v }_{ e } } -\frac { 1 }{ { u }_{ e } } =\frac { 1 }{ { f }_{ e } } \)
Where,
v_e = Image distance for the eyepiece = -d = -25 cm
u_e = Object distance for the eyepiece
\(\frac { 1 }{ { u }_{ e } } =\frac { 1 }{ { v }_{ e } } -\frac { 1 }{ { f }_{ e } } \)
\(=\frac { -1 }{ 25 } -\frac { 1 }{ 5 } =-\frac { 6 }{ 25 } \)
∴ u_e = -4.17 cm
Separation between the objective lens and the eyepiece = |u_e| + |v_o|
= 4.17 + 7.5
=11.67 cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

Area of the virtual image of each square, A = 6.25 mm²
Area of each square, A₀ = 1 mm²
Hence, the linear magnification of the object can be calculated as:
\(m=\sqrt { \frac { A }{ { A }_{ o } } } \)
\(\sqrt { \frac { 6.25 }{ 1 } } =2.5\)
But m = \(\frac{Image \ distance \ (v)}{Object \ distance \ (u)}\)
∴ v = mu
= 2.5 u ....(1)
Focal length of the magnifying glass, f = 10 cm
According to the lens formula, we have the relation:
\(\frac { 1 }{ { f } } =\frac { 1 }{ { v } } -\frac { 1 }{ { u } } \)
\(\frac { 1 }{ 10 } =\frac { 1 }{ 2.5u } -\frac { 1 }{ u } =\frac { 1 }{ u } \left( \frac { 1 }{ 2.5 } -\frac { 1 }{ 1 } \right) =\frac { 1 }{ u } \left( \frac { 1-2.5 }{ 2.5 } \right) \)
\(\therefore u=-\frac { 1.5\times 10 }{ 2.5 } =-6\)
And v = 2.5u
= 2.5 x 6 = -15 cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

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.

Height of the needle, h₁ = 4.5 cm
Object distance, u = -12 cm
Focal length of the convex mirror, f = 15 cm
Image distance = v
The value of v can be obtained using the mirror formula:
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } =\frac { 1 }{ f } -\frac { 1 }{ u } \)
\(\frac { 1 }{ 15 } +\frac { 1 }{ 12 } =\frac { 4+5 }{ 60 } =\frac { 9 }{ 60 } \)
∴ v = \(\frac { 60 }{ 9 } \) = 6.7 cm
Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.
The image size is given by the magnification formula:
\(m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } =-\frac { u }{ v } \)
\(\therefore { h }_{ 2 }=-\frac { v }{ u } \times { h }_{ 1 }\)
\(=\frac { -6.7 }{ -12 } \times 4.5=+2.5\)cm
Hence, magnification of the image, \(m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { 2.5 }{ 4.5 } =0.56\)
The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.
If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

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.

Actual depth of the pin, d = 15 cm
Apparent dept of the pin = d'
Refractive index of glass, μ = 1.5
Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
μ = \(\frac { d }{ { d }^{ ' } } \)
∴ d' = \(\frac { d }{ \mu } \)
= \(\frac{15}{1.5}\) = 10 cm
The distance at which the pin appears to be raised = d' - d
= 15 - 10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.

Focal length of the objective lens, f_o = 144 cm
Focal length of the eyepiece, f_e = 6.0 cm
The magnifying power of the telescope is given as:
m = \(\frac{f_o}{f_c}\)
= \(\frac{144}{6}\) = 24
The separation between the objective lens and the eyepiece is calculated as:
f_o + f_e
= 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Given, focal length of convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Using the formula of combination of lenses,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\)
\(\Rightarrow\) f = -60 cm
Since, the focal length of combination is negative in nature. so, the combination behaves like a diverging lens, i.e. as a concave lens.

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.

Distance between the image (screen) and the object, D = 90 cm
Distance between two locations of the convex lens, d = 20 cm
Focal length of the lens = f
Focal length is related to d and D as:
f = \(\frac { { D }^{ 2 }-{ d }^{ 2 } }{ 4D } \)
= \(\frac { { (90) }^{ 2 }-({ 20) }^{ 2 } }{ 4\times 90 } =\frac { 770 }{ 36 } =21.3\) cm
Therefore, the focal length of the convex lens is 21.39 cm.