Oswal Practice Papers CBSE Class 12 Physics Solutions (Practice Paper - 8)

Section-A

1. (d) mass of B increases

Explanation :

Negative charge means excess of electrons which increases the mass of sphere B, whereas positive charge on sphere A is given by removal of electrons.

2. (a) Electric potential

Explanation :

V = E.d = NC^–1 m

3. (b) non-ohmic conductors

Explanation :

The figure is showing I-V characteristics of non-ohmic or non-linear conductors.

4. (d)
$$\frac{\mu_0I}{2R}(1-\frac{1}{\pi})$$

Explanation :

Net magnetic field at point
$$B_{net}=B_1–B_2\\\frac{\mu_0I}{2R}-\frac{\mu_0I}{2\pi R}\\=\frac{\mu_0I}{2R}(1-\frac{1}{\pi})$$

5. (c)
$$\sqrt{\frac{2Em}{qB}}$$

6. (c) it becomes paramagnetic

Explanation :

When a ferromagnetic substance is heated to a very high temperature it looses its magnetic property Ferromagnetic substances becomes paramagnetic. This happens because of the disorderness of the electron arrangement.

7. (b)
$$\sqrt{\frac{1}{400}}s$$

Explanation :

The current takes T/4 s to reach the peak value.
In the given question,
$$XC =\frac{1}{\omega C}=\frac{1}{2\pi v C}\\XC ∝\frac{1}{v}.\\\frac{2\pi}{T}=200\pi\\⇒T=\frac{1}{1000}s\\\text{Time to reach the peak value =}\frac{1}{400}s$$

8. (d) E × B

Explanation :

The direction of propagation of electromagnetic wave is perpendicular to both electric field vector E and magnetic field vector B, i.e., in the direction of E × B. This can be seen by the diagram given below:

Here, electromagnetic wave is along the z-direction which is given by the cross product of E and B.

9. (a)
$$Q=\frac{\Delta\phi}{\Delta t}$$

Explanation :

$$\frac{\Delta\phi}{\Delta t}=\epsilon\\=IR\\⇒\Delta\phi=(I\Delta t)R\\=QR\\⇒Q=\frac{\Delta\phi}{R}$$

10. (b) decrease

Explanation :

$$\lambda=\frac{h}{mv}\\\beta=\lambda\frac{D}{d}=\frac{hD}{mvd}[\lambda=\frac{h}{mv}]\\\text{so, higher the velocity lower the fringe width.}$$

11. (c) 3 : 4

Explanation :

According to Einstein photoelectric equation
hn = h_v0 + K_max

For first metal,
hv = h
$$hv=h(\frac{3}{4}v)+K_1\\⇒K_1=\frac{hv}{4}\\\text{For second metal,}\\hv=h(\frac{3}{4}v)+K_2\\⇒K_2=\frac{hv}{3}\\\text{Thus, the ratio is}\\\frac{k_1}{k_2}=\frac{hv/4}{hv/3}=\frac{3}{4}$$

12. (c) hole concentration in p-region is more as compared to n-region.

Explanation :

The diffusion of charge carriers across a junction occurs from the region of higher concentration to the region of lower concentration. Thus, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

13. (a) Both A and R are true and R is the correct explanation of assertion.

Explanation :

In semiconductors, the energy gap between conduction band and valence band is small (≈ 1 eV). Due to temperature rise, electron in the valence band gain thermal energy and may jump across the small energy gap (to the conduction band). Thus, conductivity increases and hence resistance decreases.

14. (c) A is true, but R is false.

Explanation :

When there are few sources of light, then the result is usually called interference but if there
is a large number of them, the word diffraction is more often used.

15. (d) A is false and R is also false.

Explanation :

Photoelectric effect demonstrates particle nature of light. Number of emitted photoelectrons depends upon intensity of light.

16. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

Lenz’s Law is based on conservation of energy and induced emf opposes the cause of it i.e., change in magnetic flux.

Section-B

17. A charge oscillating with some frequency, produces an oscillating electric field in space, which in turn produces an oscillating magnetic field perpendicular to the electric field, this process goes on repeating, producing e.m. waves in space perpendicular to both the fields.

The direction of electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of the wave.

18. (i)
$$\vec{F}=q(\vec{v}×\vec{B})\\= qvBsin\theta$$

Conditions:
(a) If the angle between v and B is 90° then it will move in circular path.
(b) If the angle is other than, 0°, 90° and 180° the path will be helical.
(ii) Since the work done on the charged particle moving in the magnetic field is zero. Hence, according to work energy theorem the change in kinetic energy is zero.

19. As we know,
$$1n_2=\frac{sin(\frac{A+\Delta_m}{2})}{Sin\frac{A}{2}}\\\text{where} n_2 = \text{1.6, refractive index of glass prism,}\\n_1=\frac{4\sqrt{2}}{5}\\\text{refractive index of medium}\\\text{A = 60°, angle of prism}\\\delta=\text{angle of minimum deviation}\\\frac{1.6}{4\sqrt{2}}=sin\frac{\frac{60+\delta_m}{2}}{sin30}\\⇒\frac{0.4×5}{\sqrt{2}}=sin(\frac{{60+\delta_m}}{2})×2\\⇒sin(\frac{{60+\delta_m}}{2})=\frac{1}{\sqrt{2}}\\⇒(\frac{{60+\delta_m}}{2})=sin^{-1}(\frac{1}{\sqrt{2}})\\= sin^{–1} (sin 45°) = 45°\\⇒ δm = 30°$$

20. Given: μ = 1.33
$$λa=589 nm\\\text{We know that,}\\\mu=\frac{c}{v}\\v =\frac{c}{\mu}=\frac{3×10^8}{1.33}\\Speed,v=2.26×10^8m/s\\\text{Frequency remains same.}\\ν=\frac{c}{\lambda_a}=\frac{3×10^8}{589×10^{-9}}\\\text{Frequency, ν}=5.09×10^{14} Hz\\\text{Wavelength, λw}=\frac{\lambda_a}{\mu}\\\frac{589nm}{1.33}= 442.8 nm$$

21. The V-I characteristic of the diode is the graph drawn between the voltage, V and current I in forward bias and reverse bias of junction diode.

(i) DE is the region of negative resistance because the slope of curve in this part is negative.

(ii) BC is the region where Ohm’s law is obeyed because in this part, the current varies linearly with the voltage.

Section-C

22. At resonance, the frequency of the supply power equals the natural frequency of the given LCR circuit.
Given: Resistance, R = 20 Ω, inductance, L = 1.5 H, capacitance, C = 35 μF = 35 × 10– 6 F
AC supply voltage to the LCR circuit, V = 200 V
Impedance of the circuit is given by the relation,
$$Z=\sqrt{R^2+(X_L-X_C)}\\\text{At resonance,}X_L=X_C\\Z=R=20 \Omega\\\text{Current in the circuit can be calculated as :}\\I=\frac{V}{Z}=\frac{200}{20}=10 A\\\text{Thus, the average power transferred to the circuit is one complete cycle :}\\P=VI=200×10 = 2000 W$$

Given, Inductance (L) = 3.0 H
Capacitance (C) = 27 μF = 27 × 10– 6 F
Resistance (R) = 7.4 Ω
At resonance, angular frequency of the source for the given LCR series circuit is given by
$$wr=\frac{1}{\sqrt{LC}}\\=\frac{1}{\sqrt{3×27×10{-6}}}\\=\frac{10^3}{9}=111.11 rads^{–1}\\\text{Q-factor of the series:}\\Q=\frac{\omega_r}{R}=\frac{111.11×3}{7.4}=45.0446\\\text{To improve the sharpness of the resonance by reducing its ‘full width at half maximum’ by a factor of 2 without}\\\text{changing (wr) we need to reduce R to half i.e., resistance=}\frac{R}{2}=\frac{7.4}{2}=3.7 \Omega.$$

23. Suppose we have a conductor of length l in which current i is flowing. We need to calculate the magnetic field at a point P in vacuum.
$$If\vec{idl}\space\text{is one of the infinitely small current element, the magnetic field d}\space\vec{B}\text{at point P is given by} \\d\vec{B}∝\frac{\vec{idl}×\vec{r}}{r^3}\\d\vec{B}=\frac{\mu_0}{4\pi}\frac{\vec{idl}×\vec{r}}{r^3}\\Where\space\frac{\mu_0}{4\pi}\space\text{is a proportionality constant.}$$

Suppose there is a circular coil of radius R, carrying a current i. Let P be a point at the axis of the coil at a distance x from the centre, at which the field in required.

Consider a conducting element dl of the loop. The magnetic field due to dl is given by the Biot-Savart law,
$$d\vec{B}=\frac{\mu_0}{4\pi}\frac{\vec{i|dl}×\vec{r}|}{r^3}\\dB=\frac{\mu_0}{4\pi}\frac{idl}{(R^2+x^2)^{3/2}}$$

The direction of dB is perpendicular to the plane formed by dl and r. It has an X-component dBx and a component perpendicular to X-axis, dB⊥. When the components perpendicular to the X-axis are summed over, they cancel out and we obtain null result. Thus, only the X-component survives.
$$\text{So the resultant field}\space\vec{B}\space at P is given by\\B=\int dBsin\theta\\B=\frac{\mu_0}{4\pi}\frac{i}{r^2}\int dlsin\theta\\B=\frac{\mu_0}{4\pi}\frac{iR}{r^3}\int dl[sin\theta=\frac{R}{r}]\\But\int dl=2\pi R and r=(R^2+x^2)^{\frac{1}{2}}\\B=\frac{\mu_0}{4\pi}\frac{2\pi R^2i}{(R^2+x^2)^{\frac{3}{2}}}\\\text{If the coil has N turns, then each turn will contribute equally to B. Then,}\\B =\frac{\mu_0NiR^2}{2(R^2+x^2)^{\frac{3}{2}}}$$

24. (i) From the direction of current flow the force on each side of loop are as per fig. FAB and FDC are equal and opposite so they will cancel each other.

$$⇒ F_{AD}=\frac{\mu_0I_aI_b}{2\pi×d}×L\\=\frac{2×10^{-7}×5×2×(10×10^{-2})}{1×10^{-2}}\text{(attractive) towards left}\\= 2 × 10^{–5} N \text{(towards left)}\\F_{BC}=\frac{2×10^{-7}×5×2×(10×10^{-2})}{(1+4)×10^{-2}}\\= 4 × 10^{–6}\text{(towards right)}\\F_{net}=(F_{AD}–F_{BC})\\= 2 × 10^{–5}–0.4×10^{–5}\\=1.6×10^{– 6} N\text{(towards right)}\\\text{The direction of net force is towards the straight wire i.e., attractive.}\\\text{(ii) Torque will be zero as all the forces are in the plane of loop.}$$

25. Wavelength of incident light,
λ = 412.5 nm
= 412.5 × 10^–9m
∴ Energy of incident light.

$$E=\frac{hc}{\lambda}=\frac{6.63×10^{-34}×3×10^8}{41.25×10^{-9}}\\=4.82×10^{–19} J\\\frac{4.82×10^{–19}}{1.6×10^{–19}}\\= 3.01 eV$$

Since, the energy of incident radiation is greater than the work function of sodium and potassium, but less than that of

26. Given, energy gap of the intrinsic semiconductor (Eg) = 1.2 eV.
The temperature dependence of the intrinsic carries concentration is given by
$$n_i=n_0exp[-\frac{E_g}{2K_BT}]\\Here, k_B=\text{Boltzmann constant}=8.62×10^{–5} eV/K\\\text{(T=Temperature,} n_0=Constant)\\\text{Initial temperature}(T_1)=300K.\\\text{The intrinsic carrier-concentration at this temperature can be given by}\\n_i=n_0exp[-\frac{E_g}{2K_B×300}]...(i)\\\text{Final temperature}(T_2)=600K.\\\text{The intrinsic carrier-concentration at this temperature can be given as}\\n_i=n_0exp[-\frac{E_g}{2K_B×600}]...(ii)\\\text{The ratio between the conductivity at 600 K and at 300 K is equal to the ratio between}\\\text{the respective intrinsic carrier-concentrations at these temperature.}$$

27. (i) Net force on the charge Q/2, placed at the centre of the shell is zero.
Force on charge 2Q kept at a point A

$$F=E×2Q=\frac{1.(\frac{3Q}{2})2Q}{4\pi\epsilon_0r^2}=\frac{K×3Q^2}{r^2}\\\text{where}\space k =\frac{1}{4\pi\epsilon_0}\\\text{(ii) Electric flux through the shell,}\\\phi=\frac{Q}{2\epsilon_0}$$

28. (i) From Bohr’s model-an atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds to a certain energy level. Electron revolves is circular orbit

$$\frac{mv^2}{r}=\frac{e^2}{4\pi\epsilon_0r^2}\\\text{The motion of an electron in circular orbits is restricted in such a manner that its angular momentum}\\\text{is an integral multiple of}\frac{h}{2\pi}\\Thus, L=mvr=\frac{nh}{2\pi}\\\text{From de-Broglie hypothesis}\\λ=\frac{h}{p}=\frac{h}{mv}\\\text{And from Bohr model}\\nλ=2πr\\λ=\frac{2\pi r}{n}\\\frac{h}{mv}=\frac{2\pi r}{n}\\\frac{nh}{2\pi}= mvr=L$$

(ii) ni = 1, nf = 4.
Possible transitions are:
⇒ 4 → 3, 4 → 2, 4 → 1
or 3 → 2, 3 → 1,
or 2 → 1
Hence, six lines are possible.

$$\begin{rcases}4→1\\3→1\\2→1\end{rcases}\text{Lyman series}\\\begin{rcases}3→2\\4→2\end{rcases}\text{Balmer series}\\4→3\text{Paschen series}\\4→1\text{has smallest wavelength.}$$

(i) and (ii)

(iii)

Section-D

29.
(i) (a) Monochromatic

(a) Two
(ii) (c) Bright
(iii) (c) Interference
(iv) (b) 4 : 1,

30.
(i) (c) Nucleus
(ii) (c) Nucleus
(iii) (c) 10–15m
(iv) (c) Electrons

(d) Neutral

Section-E

31. (a) Given, capacitor are 2 μF, 3 μF and 4 μF.
(i) The combination of 2 μF and 4 μF in series with 3 μF in parallel.
$$\frac{1}{C_s}=\frac{1}{2}+\frac{1}{4}\\⇒\frac{1}{C_s}=\frac{2+1}{4}\\⇒ C_s =\frac{4}{3}\mu F\\Now, C_{equivalent} =\frac{4}{3}+{3}\mu F\\=\frac{13}{3}\mu F\\\text{So, we can get an equivalent of}\frac{13}{3}\mu F\\\text{by connecting} 2 \mu F and 4 \mu F \text{in series and} 3 \mu F \text{in parallel.}$$

(ii) Maximum capacitance can be achieved by joining the capacitors in parallel.
i.e., C_max = 2 + 3 + 4 = 9 mF
Minimum capacitance is achieved by joining the capacitors in series.
$$i.e.,\frac{1}{C_{\min}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\=\frac{6+4+3}{12}\\C_{min} =\frac{12}{13}\mu F$$

(b) Capacitance of capacitor C₁ is 100 pF.
Capacitance of capacitor C₂ is 200 pF.
Capacitance of capacitor C₃ is 200 pF.
Capacitance of capacitor C₄ is 100 pF.
Supply potential, V = 300 V
Capacitors C₂ and C₃ are connected in series. Let their equivalent capacitance be C′
$$i.e.,\frac{1}{C'}=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}\\C′=100pF\\Capacitors C_1\text{and C′ are in parallel. Let their equivalent capacitance be C′′.}\\\ C^{′′}=C^′_1+C_1\\= 100 + 100 = 200 pF\\C^{′′} and C_4 \text{are connected in series. Let their equivalent capacitance be C.}\\\frac{1}{C'}=\frac{1}{C^{"}}+\frac{1}{C_4}=\frac{1}{200}+\frac{1}{100}\\=\frac{2+1}{200}\\C=\frac{200}{3}pF\\\text{Hence, the equivalent capacitance of the circuit is}\frac{200}{3}pF.\\\text{Potential difference across}C^{′′}=V^{′′}\\\text{Potential difference across} C_4=V_4\\V^{′′}+V_4=V=300 V\\\text{Charge on}C_4\text{is given by,}\\Q_4=CV(As C_4 and C^{′′}\text{are in series then they have same charge}=Q_4)\\=\frac{200}{3}×10^{– 12}×300\\= 2×10^{– 8}C\\V_4=\frac{Q_4}{C_4}=\frac{2×10^8}{100×10^{-12}}\\\text{Voltage across} C_1 is given by,\\V_1=V–V_4=300–200=100 V\\\text{Hence, potential difference,} V_1 across C_1 is 100 V.\\\text{Charge on}C_1\text{is given by,}\\Q_1=C_1V_1\\= 100 × 10^{–12}×100=10^{– 8} C\\V2_=V_3=50 V\\\text{Therefore, charge on} C_2\text{is given by,}\\Q_2=C_2V_2\\200×10^{-12}×50=10^{-8}C\\\text{and charge on}C_3\text{given by,}\\Q_3=C_3V_3\\=200×10^{–12}×50\\= 10^{–8} C\\\text{Hence, the equivalent capacitance of the given circuit is}\frac{200}{3}pF with.\\Q_1=10– 8 C, V_1=100 V\\Q_2=10– 8 C, V_2=50 V\\Q_3=10– 8 C, V_3=50 V\\Q_4 = 2×10– 8 C, V_4=200 V\\$$

$$\text{(a) (i) Initial voltage, V1=V volts and charge stored}\\Q_1=360\mu C…(i)\\Q_1=CV_1\\\text{Changed potential,}\\V_2=V–120\\Q_2=120\mu_C\\Q_2=CV_2 …(ii)\\\text{By applying (i) divided by (ii), we get}\\\frac{Q_1}{Q_2}=\frac{CV_1}{CV_2}\\⇒\frac{360}{120}=\frac{V}{V-120}\\⇒ V=180 V\\\text{Hence,}\space C=\frac{Q_1}{V_1}=\frac{360×10^{-10}}{180}\\=2×10^{–6} F\\=2\mu F\\\text{(ii) If the voltage applied had increased by 120 V, then}\\V_3=180+120=300 V\\text{Hence charge stored in capacitor,}\\Q_3=CV_3\\=2×10^{–6}×300=600\mu C$$

(b) When the capacitors are connected in parallel.
Equivalent capacitance,
C_p = C₁ + C₂
The energy stored in the combination of the capacitors,
$$E_p=\frac{1}{2}C_pV^2\\⇒ E_p =\frac{1}{2}(C_1+C_2)(100)^2=0.25 J\\⇒(C_1+C_2)=5×10^{–5} …(i)\\\text{When the capacitors are connected in series.}\\\text{Equivalent capacitance,}\\C_s=\frac{C_1C_2}{C_1+C_2}\\\text{The energy stored in the combination of the capacitors,}\\E_s=\frac{1}{2}C_sV^2\\⇒ E_s=\frac{1}{2}\frac{C_1C_2}{C_1+C_2}(100)^2=0.045 J\\⇒=\frac{1}{2}×\frac{C_1C_2}{C_1+C_2}(100)^2=0.045 J\\⇒ C_1C_2=0.045×10^{– 4}×10^{–4}\\= 4.5×10^{–10}\\(C_1–C_2)^2=(C_1+C_2)^2–4C_1C_2\\⇒ (C_1–C_2)=\sqrt{7×10^{-10}}\\=2.64×10^{– 5}\\C_1–C_2=2.64×10^{– 5}…(ii)\\Solving (i) and (ii), we get\\C_1 = 38.2 \mu F and C_2 = 11.8 \mu F\\\text{When the capacitors are connected in parallel, the charge on each of them can be obtained as follows :}\\Q_1 = C_1V = 38.2×10^{– 6} × 100\\=38.2×10^{– 4}C\\Q_2=C_2V=11.8×10^{– 6}×100\\= 11.8×10^{– 4} C$$

32. (a) (i) Consider a parallel plate capacitor with two identical plates X and Y, each having an area of A, and separated by a distance d. Let the space between the plates be filled by a dielectric medium with its dielectric constant as K and s be the surface charge density on each of the plates.

$$\text{Surface charge density of plate 1}\space\sigma=\frac{Q}{A}\\\text{and that of plate 2 is} – \sigma.\\\text{Electric field in outer region I,}\\E =\frac{\sigma}{2\epsilon_0}-\frac{\sigma}{2\epsilon_0}=0$$

(ii) For uniform electric field, potential difference is simply the electric field multiplied by the distance between the plates, i.e.,
$$V=Ed=\frac{1}{\epsilon_0}\frac{Qd}{A}$$

(iii) Now, the capacitance of the parallel plate capacitor,
$$C =\frac{Q}{V}=\frac{Q\epsilon_0A}{Qd}=\frac{\epsilon_0A}{d}$$

(b) We know that the potential difference of the metallic sphere is given by,
$$V=\frac{Q}{4\pi\epsilon_0r}\\\text{where r is the radius of the sphere.}\\\text{Now, the potential of the metallic sphere of radius R is given by,}\\V_R=\frac{Q}{4\pi\epsilon_0R}\\V_R=\frac{\sigma(4\pi R^2)}{4\pi\epsilon_0R}\\V_R=\frac{\sigma R}{\epsilon_0}\\\text{Similarly, the potential of the metallic sphere of radius 2R is given by}\\V_{2R}=\frac{Q}{4\pi\epsilon_0 2R}\\V_{2R}=\frac{\sigma(4\pi 2R^2)}{4\pi\epsilon_02R}\\V_{2R}=\frac{\sigma×2R}{\epsilon_0}…(ii)\\\text{From the relation (i) and (ii) we know that}+ V_{2R}>V_R.$$

The charge will flow from the sphere of radius of 2R to the sphere of radius R, if the spheres are connected.

(a) Electric field on one side of a charged body is E₁ and electric field on the other side of the same body is E₂. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
$$\vec{E_1}=-\frac{\sigma}{2\epsilon_0}\hat{n}…(i)\\where,\space\hat{n}\text{= Unit vector normal to the surface at a point}\\\sigma =\text{Surface charge density at that point}\\\text{Electric field due to the other surface of the charged body,}\\\vec{E_2}=-\frac{\sigma}{2\epsilon_0}\hat{n}…(ii)\\\text{Electric field at any point due to the two surfaces,}\\\vec{E_2}-\vec{E_1}=\frac{\sigma}{2\epsilon_0}\hat{n}-(\frac{-\sigma}{2\epsilon_0}\hat{n})=\frac{\sigma}{\epsilon_0}\hat{n}\\(\vec{E_2}-\vec{E_1})\hat{n}=\frac{\sigma}{\epsilon_0}…(iii)\\\text{Since inside a closed conductor,}\\\vec{E_1}=0\\\vec{E}=\vec{E_2}=\frac{\sigma}{2\epsilon_0}\hat{n}\\\text{Therefore, the electric field just outside the conductor is}\frac{\sigma}{2\epsilon_0}\hat{n}$$

(b) When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of charged surface to the other.

33. (a)

Magnifying power : The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.
We know that,
$$\text{Magnification, m =}\frac{f_0}{f_e}=\frac{P_e}{P_o}$$

(b) (i) The lens with the smallest power or largest focal length should be used as the objective i.e., lens with power 0.5 D.
(ii) The aperature is preferred to be large so that the telescope can collect as much as light coming from the distant object as possible.

(a)
$$ \frac{sin i}{sin r}=\mu\\\text{Condition for minimum deviation:}\\1. A=180–2\alpha\\2.\frac{sin i}{sin (90-\beta)}\\r_1=r_2=r> \text{critical angle}\\r_1+r_2=180–2\alpha\\2r=180–2\alpha\\r=90–\alpha\\\beta=90–r_1\\=90–90+\alpha\\\beta=a\\\text{Condition when QR have total internal reflection:}\\∠QRC ≥ \text{critical angle for the prism}\\or ∠180°–a≥ \text{critical angle}$$

(b) Refracting telescope:
Magnifying power: The magnifying power is the ratio of the angle b subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye.

$$m = −\frac{\beta}{\alpha}=-\frac{h}{f_e}.\frac{f_o}{h}=-\frac{f_o}{f_e}$$

Limitations of refracting telescope over reflecting type telescope :
(i) Refracting telescope suffers from chromatic aberration as it uses large sized lenses.
(ii) The requirement of big lenses tend to be very heavy and therefore, difficult to make and support by their edges.