Geometric Optics- Multiple Choice Questions involving Refraction at Plane Surfaces

I am neither especially clever nor especially gifted. I am only very, very curious.

– Albert Einstein

The following question at the first glance may appear to be a difficult one to many of you; but, you will realise how easy it is when you apply basic points you studied in geometric optics:

A glass jar has the plane inner surface PQ of its bottom silvered and contains water (of refractive index n = 4/3) column of height t = 6 cm. A small light emitting diode (LED) is arranged at O at a height d = 8 cm from the water surface AB (Fig.). The silvered bottom of the jar acts as a plane mirror. At what distance from the free surface (AB) of water will this plane mirror form the image of the light emitting diode?

(a) 11 cm

(b) 14 cm

(c) 17 cm

(d) 18 cm

(e) 20 cm

When you look into the plane mirror (silvered surface) PQ from the position O of the LED, the plane mirror will appear to be located at P₁Q₁ (fig.) at a distance t/n from the free surface of water (because of normal refraction at the water surface). The distance of the LED from this refracted image P₁Q₁ of the plane mirror is therefore equal to (d + t/n) as shown in the adjoining figure.

The image of the LED must be formed at O₁ which is at the same distance (d + t/n) from the effective plane mirror P₁Q₁.

As is clear from the adjoining figure, the distance of the image O₁ from the free surface (AB) of water is (d + t/n) + t/n which is equal to (d + 2t/n) = 8 + 2×6/(4/3) = 17 cm.

The following question appeared in EAMCET (Engineering) 2003 question paper:

One of the refracting surfaces of a prism of refractive index √2 is silvered. The angle of the prism is equal to the critical angle of a medium of refractive index 2. A ray of light incident on the unsilvered surface passes through the prism and retraces its path after reflection at the silvered face. Then the angle of incidence on the unsilvered surface is

(a) 0º

(b) 30º

(c) 45º

(d) 60º

The angle A of the prism (as mentioned in the question) is given by n = 1/sin A where n = 2.

[Remember n = 1/sin C where n is the refractive index and C is the critical angle].

Therefore, sin A = ½ so that A = 30º

Since the ray retraces its path after reflection at the silvered face, it is incident normally at the silvered face (at the point N in the figure). With reference to the figure, angle QNA in the triangle QNA is 90º.

Since the angle A is 30º it follows that angle AQN = 60º so that the angle of refraction (r) at Q is 30º.

The angle of incidence (i) at the unsilvered face is given by

n = sin i/sin r from which sin i = n sin r = √2 sin 30º.

This gives sin i =1/√2 so that i = 45º.

You may search for ‘optics’ using the ‘search blog’ facility at the top left of this page to find all related posts on this site.

A useful post on the equations to be remembered in Geometric Optics can be found here.

Kerala Engineering Entrance 2008 and other Multiple Choice Questions (MCQ) on Work and Energy

Setting an example is not the main means of influencing others; it is the only means.

– Albert Einstein

Let us discuss a few multiple choice questions on work and energy.

(1) A small sphere of mass 20 g is projected vertically up with a velocity of 10 ms^–1. If air resistance is negligible, what is the total work done by gravity during the upward trip of the sphere?

(a) 10 J

(b) –10 J

(c) – 1 J

(d) 1 J

(e) 9.8 J

When the sphere rises up its kinetic energy goes on decreasing because of the work done by the gravitational force against the motion of the sphere. (The gravitational potential energy of the sphere goes on increasing by an equal amount). The work done by gravity is evidently negative. When the sphere reaches the maximum height the entire kinetic energy gets converted into gravitational potential energy. The total work done by gravity during the the upward trip of the sphere is numerically equal to the initial kinetic energy (½ mv²) of the sphere but its sign is negative.

Therefore, the answer is – ½ mv² = – ½ ×0.020×10² = –1 J.

(2) A world class athlete covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

(a) 2 × 10⁵ J to 5 × 10⁵ J

(b) 2 × 10⁴ J to 5 × 10⁴ J

(c) 2 × 10³ J to 5 × 10³ J

(d) 200 J to 500 J

(e) 20 J to 50 J

This question has appeared in various entrance tests with slight differences in the wording and in the options. Last year it appeared in the All India Engineering Entrance Examination.

In the 100 m dash the velocity v is almost constant so that we have

v = 100 m/10 s = 10 ms^–1

We may take the mass of the athlete to be 70 kg to 80 kg. Let us use 80 kg.

His kinetic energy will be ½ mv² = ½ ×80×10² = 4000 J so that the correct option is (c).

Questions (3) and (4) appeared in Kerala Engineering Entrance Examination 2008 question paper.

(3) Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times t_A and t_B, then the ratio t_A/ t_B is

(a) ½

(b) 2

(c) 2/5

(d) 5/6

(e) 1/5

The concept of impulse will be very useful here. The impulse received by A and B in the times t_A and t_B are respectively F t_A and F t_B where F is the force acting on them (4 kg wt. here). But impulse is the change in momentum so that the momenta acquired by A and B are F t_A and F t_B.

Since the kinetic energy is p²/2m where p is the momentum, we have

(F t_A)²/2m_A = (F t_B)²/2m_B

Therefore, t_A/ t_B = √( m_A/ m_B) = √(20/5) = 2.

(4) A particle acted upon by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces in SI units is

(a) 20

(b) 40

(c) 50

(d) 30

(e) 35

The resutant force (F) on the particle is the sum of the forces given by

F = (4i + j – 3k) + (3i + j – k) = 7i + 2j – 4k.

The displacement (s) of the particle is given by

s = (5i + 4j + k) – (i + 2j + 3k) = (4i + 2j – 2k)

The work done (W) is given by

W = F.s = (7i + 2j – 4k) . (4i + 2j – 2k) = 28 + 4 + 8 = 40.

You will find similar useful multiple choice questions (with solution) at AP Physics Resources.

KEAM 2009 Rescheduled

The Commissioner for Entrance Examinations, Govt. of Kerala, has notified that the dates of the Entrance Examinations for Admission to Medical/ Agriculture/ Veterinary/ Engineering Degree Courses 2009 (KEAM 2009), Kerala have been changed (for the second time) as follows:

Modified Dates of Exam:

Engineering Stream:

25.05.2009 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.

26.05.2009 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Stream:

27.05.2009 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.

28.05.2009 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

For more details visit the site http://www.cee-kerala.org/ where you will find information regarding NATA (National Aptitude Test in Architecture) required for those who apply for Architecture

Angular Momentum- Two Multiple Choice Questions

How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though some times sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our happiness depends.

– Albert Einstein

(1) A thin uniform wooden rod AB of length L and mass M is hinged without friction at the end B. A small block of clay of mass m moving horizontally with velocity v srikes the end B of the rod and gets stuck to it. The angular velocity of the system about A just after the collision is

(a) mv/(mL + ML)

(b) 3mv/(mL + ML)

(c) (mv+ ML)/ 3mL

(d) mv/(3mL + ML)

(e) 3mv/(3mL + ML)

The angular momentum of the system about the point A just before collision is the angular momentum of the block of clay which is equal to mvL. (Note that the perpendicular distance of the line of action of the linear momentum (mv) of the block of clay from the point A is L).

The angular momentum of the system about the point A just after the collision is Iω where I is the total moment of inertia of the rod and clay and ω is the angular velocity of the system immediately after the collision.

We have I = ML²/3 + mL²

Equating the angular momenta before and after collision, we have

mvL = (ML²/3 + mL²) ω

Therefore ω = 3mv/(3mL + ML).

(2) A particle is projected at an angle of 60º with the horizontal with linear momentum of magnitude p. The horizontal range of this projectile is R. Just before the projectile strikes the ground at A, what is the magnitude of its angular momentum about an axis perpendicular the plane of motion and passing through the point of projection? Neglect air resistance.

(a) pR

(b) pR/2

(c) (√3) pR /2

(d) (√3) pR

(e) Zero

This is a very simple question. The magnitude of the linear momentum of the projectile just before it strikes the ground will be equal to p. The magnitude of the angular momentum at the moment will be p×Lever arm = p×ON = p×R sin 60º = (√3) pR /2.

IIT-JEE 2008 Linked Comprehension Type Multiple Choice Questions Inviolving Thermal Physics and Properties of Fluids

Today we will discuss three linked comprehension type multiple choice questions which appeared in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

A small spherical monoatomic ideal gas bubble (γ = 5/3) is trapped inside a liquid of density ρ_ℓ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T₀, the height of the liquid is H and the atmospheric pressure is P₀ (Neglect surface tension).

1. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it:

(A) Only the force of gravity

(B) The force due to gravity and the force due to the pressure of the liquid

(C) The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid

(D) The force due to gravity and the force due to viscosity of the liquid

The force due to the pressure of the liquid is the buoyancy force. Therefore, besides the buoyancy force the forces acting on the bubble are the force due to gravity and the force due to viscosity of the liquid [Option (D)].

2. When the gas bubble is at height y from the bottom, its temperature is

(A) T₀[(P₀ + ρ_ℓ gH) /(P₀ + ρ_ℓ gy)]^2/5

(B) T₀[{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ gH)]^2/5

(C) T₀[(P₀ + ρ_ℓ gH) /(P₀ + ρ_ℓ gy)]^3/5

(D) T₀[{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ gH)]^3/5

Since there is no heat exchange, the process is adiabatic for which

T P^{[(1- γ)/ γ]} = constant where γ is the ratio of specific heats.

[Many of you might be remembering this as T ^γ P^{(1- γ)} = constant].

At the bottom pressure = P₀ + ρ_ℓ gH and temperature = T₀

At height y from the bottom pressure = P₀ + ρ_ℓ g(H –y) and temperature = T (let us say)

Therefore, T₀(P₀ + ρ_ℓ gH)^–2/5 = T[P₀ + ρ_ℓ g(H –y)]^–2/5 so that

T = T₀[(P₀ + ρ_ℓ gH) /{P₀ + ρ_ℓ g(H –y)}]^–2/5

Or, T = T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5

So option (B) is correct.

3. The buoyancy force on the gas bubble is (Assume R is the universal gas constant)

(A) ρ_ℓ nRgT₀ [(P₀ + ρ_ℓ gH) ^2/5/(P₀ + ρ_ℓ gy) ^7/5]

(B) ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^2/5{P₀ + ρ_ℓ g(H –y)}^3/5]

(C) ρ_ℓ nRgT₀ [(P₀ + ρ_ℓ gH) ^3/5/(P₀ + ρ_ℓ gy) ^8/5]

(D) ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^3/5{P₀ + ρ_ℓ g(H –y)}^2/5]

Force of buoyancy, F= Vρ_ℓ g

But V = nRT/P so that F = nRTρ_ℓ g /P

Here P = P₀ + ρ_ℓ g(H –y) and T = T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5

Substituting, F = nRρ_ℓ g T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5/ [P₀ + ρ_ℓ g(H –y)]

Or, F = ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^2/5{P₀ + ρ_ℓ g(H –y)}^3/5]

So option (B) is correct.