A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eye-balls get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened. When the eye-lens loses its ability of accommodation, the defect is called presbyopia.

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.

Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\)  Output frequency = 2 x 50 = 100 Hz.

Reflecting telescope consists of concave mirror of large aperture and large focal length (objective). A plane mirror is placed between the concave mirror and its focus. A small convex lens works as eyepiece.

Advantages of Reflecting Telescope over Refracting Telescope
For a reflecting telescope, the mirror affords several advantages over the objective lens in astronomical telescope
(i) A mirror is easier to produce with a larger diameter, so that it can intercept rays crossing a larger area and direct them to the eyepiece.
(ii) The mirror can be made parabolic to reduce spherical aberration. Aberration is further reduced because passage through one layer of glass (the objective lens) is eliminated.

Given, V = 0.4 V
(i) d = 4 \(\times\) 10^-7 m, E = ?
Electric field, \(E=\frac{V}{d}=\frac{0.4}{4 \times 10^{-7}}\)
\(=1 \times 10^6 \mathrm{~V} / \mathrm{m}\)
(ii) v₁ = 4 \(\times\) 10⁵ m/s, v₂ = ?
Let v₁ be the speed of electron when it enters the depletion layer and v₂ be the speed when it comes out of the depletion layer.
According to principle of conservation of energy, KE before entering the depletion layer = loss in PE + KE after crossing the depletion layer
\(\begin{aligned} \Rightarrow \quad \frac{1}{2} m v_1^2=e \times V+\frac{1}{2} m v_2^2 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{1}{2} \times 9.1 \times 10^{-31} \times\left(4 \times 10^5\right)^2 \end{aligned}\)
\(\begin{aligned} =1.6 \times 10^{-19} \times 0.4 +\frac{1}{2} \times 9.1 \times 10^{-31} \times v_2^2 \end{aligned}\)
\(\therefore\) v₂ = 1.39 \(\times\) 10⁵ m/s

Given, cut-off voltage, V₀ = 1.5 V
Maximum kinetic energy is given by,
\(\begin{aligned} \mathrm{KE}_{\text {max }} & =e V_0=1.5 \space \mathrm{eV}=1.5 \times 1.6 \times 10^{-19} \end{aligned}\)
\(=2.4 \times 10^{-19} \mathrm{~J}\)

Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be
I = I₀ cos ² \(\theta\) 
where \(\theta\) is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be ( \(\pi\) / 2 –\(\theta\) ). Hence the intensity of light emerging from P₃ will be
\(I=I_{0} \cos ^{2} \theta \cos ^{2}\left(\frac{\pi}{2}-\theta\right)\)
\(=I_{0} \cos ^{2} \theta \sin ^{2} \theta=\left(I_{0} / 4\right) \sin ^{2} 2 \theta\)
Therefore, the transmitted intensity will be maximum when \(\theta\) = \(\pi\) / 4.

The image of the object in a plane mirror is as far behind the mirror as the object is in front of it. Therefore, the image of the boy comes near the mirror through the distance equal to that moved by the boy towards the plane mirror. Hence, the image of the boy will approach him with double his speed, i.e. with 4 m/s.

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.

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.

\( \text { As } O Q>O P \\ \therefore v_0^{\prime}>v_0 \)
So, the threshold frequency of metal A is greater than metal B.

\(V_0\) When the distance is doubled, the intensity of incident light becomes one-fourth, but the frequency of incident light remains unchanged. The value of cutoff potential of a material only depends upon the frequency of the incident light and is independent of the intensity of incident light.

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.

(a) Refractive index of glass, μ = 1.5
Speed of light, c = 3 x 10⁸ m/s
Speed of light in glass is given by the relation
\(v=\frac{c}{\mu}\)
\(=\frac{3 \times 10^{8}}{1.5}=2 \times 10^{8} \mathrm{~m} / \mathrm{s}\)
Hence, the speed of light in glass is 2 x 10⁸ m/s.
(b) The speed of light in glass is not independent of the colour of light
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

(i) Energy = eV= hv
or \(v=\frac{e V}{h}=\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}}\)
= 7.24 \(\times\) 10¹⁸ Hz
(ii) As, \(c=v \lambda\)
\(\therefore\) Wavelength, \(\lambda=\frac{c}{v}=\frac{3 \times 10^8}{7.24 \times 10^{18}}=0.0414 \mathrm{~nm}\)

(i) (d): Refractive index ofa medium depends upon nature and temperature of the medium, wavelength of light.
(ii) (a): Here \(v=5 \times 10^{14} \mathrm{~Hz} ; \lambda=450 \times 10^{-9} \mathrm{~m}\)
\(c=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\)
Refractive index of the liquid,
\(\mu=\frac{c}{v}=\frac{c}{v \lambda}=\frac{3 \times 10^{8}}{5 \times 10^{14} \times 450 \times 10^{-9}} \)
\(\mu=1.33\)
(iii) (b): Here i = 60° ; \(\mu\)= 1.5
By snell's law, \(\mu=\frac{\sin i}{\sin r}\)
\(\sin r=\frac{\sin i}{\mu}=\frac{\sin 60^{\circ}}{1.5}=\frac{0.866}{1.5} \)
\(\sin r=0.5773 \text { or } r=\sin ^{-1}(0.58)\)
(iv) (c): As object is at the centre of the sphere, the image must be at the centre only.
\(\therefore\) Distance of virtual image from centre of sphere = 6cm.
(v) (c): Speed of light in second medium is different than that in first medium.