Geometrical Optics- Questions (MCQ) on Refraction at Prisms

Questions involving the refraction produced by prisms often find place in Medical, Engineering and other Degree Entrance Exam question papers. Here are two questions in this section:

(1) The face AC of a glass prism of angle 30º is silvered. A ray of light incident at an angle of 60º on face AB retraces its path on getting reflected from the silvered face AC. If the face AC is not silvered, the deviation that can be produced by the prism will be

(a) 0º

(b) 30º

(c) 45º

(d) 60º

(e) 90º

The deviation (d) produced by a prism is given by

d = i₁ + i₂ – A where i₁ and i₂ are respectively the angle of incidence and the angle of emergence and A is the angle of the prism. Here we have i₁ = 60º, i₂ = 0º (since the ray falling normally will proceed undeviated from the face AC if it is not silvered) and A = 60º.

Therefore, d = 30º.

(2) In the above question, what is the refractive index of the material of the prism?

(a) 1.732

(b) 1.652

(c) 1.667

(d) 1.5

(e) 1.414

In the triangle ADN angle AND is 60º since angle DAN = 30º and angle DNA = 90º. Therefore, the angle of refraction at D is 30º. The refractive index of the material of the lens (n) is given by

n = sin i₁/ sinr₁ = sin 60º/ sin 30º = √3 = 1.732

(3) Two thin (small angled) prisms are combined to produce dispersion without deviation. One prism has angle 5º and refractive index 1.56. If the other prism has refractive index 1.7, what is its angle?

(a) 3º

(b) 4º

(c) 5º

(d) 6º

(e) 7º

Since the deviation (d) produced by a small angled prism of angle A and refractive index n is given by

d = (n – 1)A, the condition for dispersion without deviation on combining two prisms of angles A₁ and A₂ with refractive indices n₁ and n₂ respectively is

(n₁ – 1)A₁ = (n₂ – 1)A₂

Therefore, 0.56×5 = 0.7×A₂ so that A₂ = 4º

On this site you will find many questions (with solution) on refraction at plane surfaces as well as at curved surfaces. To access all of them type in ‘refraction’ in the search box at the top left side of this page and click on the adjacent ‘search blog’ box.

All India Pre-Medical / Pre-Dental Entrance Examination -2009 (AIPMT 2009)

Central Board of Secondary Education, Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination -2009 as per the following schedule for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments, Municipal or other local authorities in India except in the States of Andhra Pradesh and Jammu & Kashmir:-

1. Preliminary Examination - 5th April, 2009 (Sunday)

2. Final Examination - 10th May, 2009 (Sunday)

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination ither offline or online as explained below:

Offline

Offline submission of Application Form may be made using the prescribed application form. The Information Bulletin and Application Form costing Rs.600/- (including Rs.100/- as counselling fee) for General & OBC Category Candidates and Rs.350/- (including Rs.100/- as counselling fee) for SC/ST Category Candidates inclusive of counseling fee can be obtained against cash payment from 22-10-2008 to 01-12-2008 from any of the branches of Canara Bank/ Regional Offices of the CBSE. Find details at http://www.aipmt.nic.in/.

Online

Online submission of application may be made by accessing the Board’s website http://www.aipmt.nic.in/. from 22-10-2008 (10.00 A.M.) to 01-12-2008 (5.00 P.M.). Candidates are required to take a print of the Online Application after successful submission of data. The print out of the computer generated application, complete in all respect as applicable for Offline submission should be sent to the Deputy Secretary (AIPMT), Central Board of Secondary Education, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post only. Fee of Rs.600/-(including Rs.100/- as counseling fee) for General and OBC Category Candidates and Rs.350/- (including Rs.100/- as counseling fee) for SC/ST category candidates may be remitted in the following ways :

1. By credit card or

2. Through Demand Draft in favour of the Secretary, Central Board of Secondary Education, Delhi drawn on any Nationalized Bank payable at Delhi. Instructions for Online submission of Application Form is available on the website http://www.aipmt.nic.in/. Application Form along with original Demand Draft should reach the Board on or before 04-12-2008 which is the last date stipulated.

Visit the web site http://www.aipmt.nic.in/. for all details and information updates.

Questions (MCQ) on Newton’s Laws

The following simple questions may prompt you to pick out the wrong option if you are in a hurry. So, be cautious and don’t overlook basic points. Here are the questions:

(1) Two identical frictionless pulleys carry the same mass 2m at the left ends of the light inextensible strings passing over them. The right end of the string carries a mass 3m in the case of arrangement (i) where as a force of 3mg is applied in the case of arrangement (ii) as shown in the adjoining figure. The ratio of the acceleration of mass 2m in case (i) to the acceleration of mass 2m in case (ii) is

(a) 2:3

(b) 1:1

(c) 5:3

(d) 3:5

(e) 2:5

In both cases the net driving force is 3mg – 2mg = mg. But in case (i) the total mass moved is 5m where as in case (ii) the total mass moved is 2m. The acceleration in case (i) is mg/5m = g/5 where as the acceleration in case (ii) is mg/2m = g/2.

The ratio of accelerations = (g/5)/ (g/2) = 2/5 [Option (e)].

(2) An object of mass 4 kg moving along a horizontal surface with an initial velocity of 2 ms^–1 comes to rest after 4 seconds. If you want to keep it moving with the velocity of 2 ms^–1, the force required is

(a) zero

(b) 1 N

(c) 2 N

(d) 4 N

(e) 8 N

The acceleration ‘a’ of the body is given by

0 = 2 + a×4, on using the equation, v_t = v₀ + at

Therefore, a = – 0.5 ms^–2

The body is retarded (as indicated by the negative sign) because of forces opposing the motion. The opposing force has magnitude ma = 4×0.5 = 2 N. Therefore, a force of 2 N has to be applied opposite to the opposing forces to keep the body moving with the velocity of 2 ms^–1.

You will find some useful multiple choice questions (with solution) in this section here as well as here