Optics-Questions on Interference

The following MCQ appeared in Kerala Engineering Entrance 2005 test paper:

In the Young’s double slit experiment, the intensity of the central maximum is observed to be I₀. If one of the slits is covered, the intensity at the central maximum will become

(a) I₀/2 (b) I₀/√2 (c) I₀/4 (d) I₀ (e) I₀²

If the resultant amplitude (due to the two interfering waves) at the central maximum is ‘a’. we can write

I₀ α a², since the intensity is proportional to the square of the amplitude.

When one of the slits is covered, the amplitude is reduced to a/2. If ‘I’ is the intensity at the position of the central maximum now, we can write

I α (a/2)².

From the above, we obtain I = I₀/4 [Option (c)].

Now, consider the following question:

In a double slit interference pattern, the intensity at the centre of a bright fringe is I. The intensity at a point one quarter of the distance to the next bright fringe is

(a) I/2 (b) I/4 (c) I/8 (d) I (a) zero

At the centre of a bright fringe the waves arrive in phase. You may imagine that the photons starting from the two slits are in the same state of vibration when they reach the position of the centre of a bright fringe and that is why their amplitudes get added to produce maximum intensity. At the centre of the next bright fringe, the photons will have an extra phase difference of 2π, but this too is ‘in phase’ condition (for the same state of vibration).

At a point one quarter of the distance to the next bright fringe, the phase difference between the interfering photons will be 2π/4 = π/2.

If ‘a’ is the amplitude of each interfering wave, the resultant amplitude at the centre of a bright fringe is 2a and the intensity I is given by

I α 4a²

At a point one quarter of the distance to the next bright fringe, the amplitudes are added with a phase difference of π/2 and the resultant amplitude is √(a² + a²) = √2 a. The intensity (I') in this case is given by

I' α 2a²

From the above expressions, we obtain I' = I/2 [Option (a)].

[Note that the intensity (I) produced by two interfering waves of the same amplitude ‘a’ is given by I α 4a²cos²(δ/2) where ‘δ’ is the phase difference].

You will find more multiple choice questions with solution in this section at physicsplus: Multiple Choice Questions on Wave Optics and at physicsplus: Questions on Polarisation

Questions from Thermodynamics

(1) In the cyclic process shown in the PV diagram for an ideal gas, the net work done by the gas during one cycle is
(a) 24P₀ V₀ (b) –24P₀V₀ (c) – 6πP₀V₀ (d) 6πP₀V₀ (e) 4πP₀V₀

The area enclosed by the closed curve gives the work done. Since the cycle of operations is clockwise (in the PV diagram with P on the Y-axis), the net work done by the gas is positive

Therefore net work done by the system during one cycle = Area of circle.
Since the quantities on the two axes are different, the shape is not rally a circle. The scales of the quantities could have been changed to make it an ellipse. So, treat it as an ellipse and calculate the area as πab where ‘a’ and ‘b’ are the semi-major and semi-minor axes respectively.
So, work done = π[(5P₀ –2P₀)/2] [(10V₀ –2V₀)/2] = 6πP₀V₀
[Note that if the arrow in the diagram shows the processes in the anticlockwise direction, the net work done by the system will be negative].
(2) A refrigerator removes 1000 calories of heat from the ice trays. The coefficient of performance of the refrigerator is 10. Then the work done by the compressor motor is at least
(a) 240 J (b) 420 J (c) 4200 J (d) 42000 J (e) 0.042 J
Coefficient of performance, β = Q2/w where Q₂ is the quantity of heat removed from the cold body (inside the refrigerator) and ‘w’ is the work needed to transfer this heat to the hot body (outside).
Therefore w = Q/β = 1000/10 = 100 calories = 420 J, approximately. This is the minimum work to be done by the compressor motor.
(3) An engine takes in 10000 J of heat and rejects 8000 J while operating between temperatures of 900 K and 600 K. The actual efficiency is
(a) 1/5 (b) 2/5 (c) 1/3 (d) 2/3 (e)4/5
This simple question appeared in the Kerala Engineering Entrance 2000 test paper. The actual efficiency is given by η = (Q₁ –Q₂)/Q₁. Note that this will be less than (T₁–T₂)/T₁ in practical heat engines. The efficiency given by the two expressions will be equal in the case of the ideal, perfectly reversible Carnot engine only.
Actual efficiency = (10000–8000)/10000 = 1/5.

Two Questions (MCQ) on Telescopes

nvppfThe following question appeared in MPPMT 2000 question paper:
The diameter of the objective of a telescope is ‘a’, its magnifying power is ‘m’ and the wave length of light is λ. The resolving power of the telescope is
(a) a/1.22 λ (b) λm/1.22a (c) 1.22a/ λ (d) a/1.22 λm
The magnifying power given in the problem is just a distraction. The limit of resolution of a telescope, which is the minimum angular separation between two objects that can be resolved is given by dθ = 1.22λ/a. The resolving power is the reciprocal of the limit of resolution and is equal to a/1.22 λ. [Option (a)].
Objectives and eye pieces of telescopes are made of achromatic combination of lenses to make them devoid of the defect of chromatic aberration. Consider the following MCQ:
A telescope objective is an achromatic doublet made of crown glass and flint glass lenses. The proper choice for the achromatic combination is
(a) divergent lens of flint glass and convergent lens of crown glass
(b) convergent lens of flint glass and divergent lens of crown glass
(c) both convergent (d) both divergent
(e) as given in options (a) or (b) or (c)
You might be remembering the condition for an achromatic doublet: ω₁ /f₁+ ω₂/f₂ = 0, from which ω₁/f₁= – ω₂/f₂.
Since the dispersive power ‘ω’ is a positive quantity, the negative sign shows that one lens must be diverging (negative focal length) and the other converging. In a telescope, the objective has to be converging. Therefore, the converging lens should have smaller focal length (larger power) so that when combined with the diverging lens, the combination will still be converging. Crown glass has smaller dispersive power and hence the converging lens should be made of crown glass to satisfy the above condition of achromatism [Option (a)].

You will find more multiple choice questions (with solution) on telescopes, microscopes and the defects of the eye at physicsplus: Questions on Optical Instruments

Kerala Government Medical and Engineering Entrance Examinations-2007

The Commissioner of Entrance Examinations (Govt. of Kerala) has invited applications for the Entrance Examinations for admission to the following Professional Degree courses in Kerala for 2007-08:
(a) Medical (i) MBBS (ii) BDS (iii) BPharm (iv) BSc (Nursing) (v) BSc (MLT) (vi) BAMS (vii) BHMS (viii) BSMS (Siddha) (ix) BSc–Nursing (Ayurveda) and (x) BPharm (Ayurveda) .
(b) Agriculture (i) BSc (Agriculture) (ii) BFSc (Fisheries) (iii) BSc (Forestry)
(c) Veterinary BVSc & AH
(d) Engineering B.Tech [including BTech (Agricultural Engg. / BTech (Dairy Sc. & Tech) courses under the Kerala Agr iculture University]
(e) Architecture B.Arch
Time Table for the Examinations:
The Entrance Examinations will be held in all the District Centres in Kerala, New Delhi and Dubai (UAE), on the dates mentioned below as per Indian Standard Time.
Engineering Entrance Examination (For Engineering courses except Architecture):
23-04-2007 Monday 10.00 A.M. to 12.30 PM Paper-I: Physics & Chemistry.
24-04-2007 Tuesday 10.00 A.M. to 12.30 PM Paper-II: Mathematics.
Medical Entrance Examination [For Medical (including B.Pharm), Agriculture and Veterinary Courses]:
25-04-2007 Wednesday 10.00 A.M. to 12.30 PM Paper-I: Chemistry & Physics.
26-04-2007 Thursday 10.00 A.M. to 12.30 PM Paper-II: Biology.
Application Forms:
The application form and Prospectus will be sold from 07-02-2007 to 05-03-2007 through selected Canara Bank branches in Kerala and outside the State.
Cost of Application form: General candidates: Rs. 700/- ; SC/ST candidates : Rs. 350/- ( Candidates opting Dubai centre should enclose a bank draft for Rs 7000/- along with the application)
Last date and time for receipt of filled in Application Forms: The filled in Application Form along with the OMR DATA SHEET and other relevant documents to be submitted with the Application form is to be sent in the printed envelope bearing the address of the Commissioner for Entrance Examinations supplied along with the application form so as to reach him before 5 p.m. on 07.03 200 7, by Hand Delivery / Registered Post/ Speed Post.
Visit the site www.cee-kerala.org for complete details regarding the eligibility for applying for these examinations, sale centres of application forms, provision for downloading the application form in the case of certain categories etc. Make it a habit to visit the site to be informed of the information updates in this connection.

Questions on Electrostatics

The following question which appeared in IIT 1997 Entrance test paper checks whether you have a thorough understanding of the basic relation between the electric field and potential:
A non-conducting ring of radius 0.5m carries a total charge of 1.11×10^–10 C distributed non uniformly on its circumference producing an electric field E everywhere in space. The value of the line integral ∫ –E.dl between the limits l = ∞ to l = 0 ( l = 0 being the centre of the ring) in volts is
(a) +2 (b) –1 (c) –8 (d) zero
The line integral ∫ –E.dl between the limits l = ∞ and l = 0 gives the work done in bringing unit positive charge from infinity to the centre of the ring and therefore is equal to the electric potential at the centre of the ring. If the total charge on the ring is Q coulomb, the potential at the centre is V = (1/4πε₀)×Q/r where ‘r’ is the radius of the ring. Therefore, V = 9×10⁹ ×1.11×10^–10/0.5 = 2, very nearly. So, the correct option is (a).
Now, consider the following MCQ:
A dielectric slab (dielectric constant = K) of thickness ‘t’ is placed between the plates of a parallel plate air capacitor. If the capacitance of the capacitor is to be restored to the original value, the separation between the plates is to be increased by
(a) kt (b) k/t (c) t/k (d) t –(k/t) (e) t –(t/k)
When a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space are respectively q/ε₀A and q/Kε₀A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε₀A)(d–t)+(q/Kε₀A)t = (q/ε₀A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε₀A/[d–t+(t/K)] = ε₀A/[d–(t – t/K)].
Since the capacitance of the capacitor with air filling the entire space between the plates is ε₀A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).
The following MCQ appeared in Kerala Engineering Entrance 2003 question paper:
A parallel plate capacitor has a capacitance of 100 pF when the plates of the capacitor are separated by a distance of ‘t’. Then a metallic foil of thickness t/3 is introduced between the plates. The capacitance will then become
(a) 100 pF (b) (3/2)100 pF (c) (2/3)100 pF (d) (1/3)100 pF (e) (1/2)100 pF
As shown in the previous discussion, the capacitance of a parallel plate capacitor with a dielectric slab of thickness ‘t’ between the plates separated by a distance ‘d’ is given by,C = ε₀A/[d–t+(t/K)]. In the present problem, ‘d’ is to be replaced by ‘t’ and ‘t’ is to be replaced by t/3. Further, the dielectric constant K is to be replaced by ∞ since the dielectric constant of a conductor is infinite. The capacitance therefore becomes ε₀A/(t–t/3) = (3/2) ε₀A/t = (3/2)100 pF (since the original capacitance with air alone as the dielectric is ε₀A/t = 100 pF).

You will find more multiple choice questions with solution at physicsplus: Questions on Electrostatics