Wednesday, November 22, 2006

Questions (MCQ) on Refraction at Spherical Surfaces

Some of you might be scared of questions on reflection and refraction at spherical surfaces. This unnecessary fear is due to your confusion about the inevitable sign convention used while discussing spherical mirrors and lenses.
The sign convention used widely in Optics is the Cartesian sign convention. The ray incident on the curved surface is to be considered as proceeding in the positive X-direction and you have to measure all distances from the pole, which is supposed to be the origin.
Often you may be given a ray diagram in which the incident ray may be proceeding in the negative X-direction. To avoid confusion, imagine that the direction of the incident ray is still in the positive X-direction. You can even redraw the diagram to make the incident ray proceed from left to right if you want. The signs given to the distances are as in the Cartesian coordinate system: Distances measured from the pole towards right are positive and distances measured from the pole towards left are negative. Similarly, distances measured (from the pole) upwards are positive and those measured downwards are negative. Normally you will encounter problems involving leftward and rightward measurements.
While solving problems, apply signs to all known quantities in accordance with the sign convention. Leave the unknown quantities as they are in the formulae. You will be able to arrive at conclusions by interpreting the sign of the unknown quantity you arrive at finally as the answer. For instance, if the distance of an image is obtained as negative, you will immediately understand that the image is on the same side of the curved surface as the object is. If the focal length of a lens is obtained as negative, you will understand that it is a concave lens. If the focal length or the radius of curvature of a mirror is obtained as positive, you will understand that it is a convex mirror.
Here are the essential formulae you have to remember in connection with refraction at spherical surfaces:
(1) Law of distances in the case of refraction at a spherical interface between two media of refractive indices
n1 and n2:
n2 /v – n1/u = (n2 – n1)/R where ‘v’ is the image distance, ‘u’ is the object distance and ‘R’ is the radius of curvature of the refracting surface.
(2) Lens maker’s equation: 1/f = (n2/n1 – 1)(1/R1 – 1/R2) where ‘f’ is the focal length of the lens, R1 and R2 are its radii of curvature, n2 is its refractive index and
n1 is the refractive index of the medium in which the lens is placed.
If the lens is placed in air or vacuum, lens maker’s equation becomes 1/f = (n – 1)(1/R1 – 1/R2) where 'n' is the refractive index of the lens. [It is better that you remember the general Lens maker’s equation since you will encounter questions in which the lens is placed in liquids such as water].
(3) Law of distances (Lens formula): 1/f = 1/v - 1/u where ‘f’ is the focal length of the lens, ‘v’ is the image distance and ‘u’ is the object distance.
(4) Focal length (F) of the combination of two or more lenses in contact:
1/F = 1/f1 + 1/f2 + 1/f3 + …etc. where f1,f2,f3...etc. are the focal lengths of the individual lenses. [ Note that this is a power equation since the reciprocal of focal length is the power (focal power) of the lens].
(5) Focal length of the combination of two lenses separated by a distance:
1/F = 1/f1 + 1/f2 – d/f1f2 where ‘d’ is the separation.
(6) Formula for calculating the focal length by displacement method (conjugate position method): f = (D2 – d2)/4D where D is the distance between the object and the screen (image) and ‘d’ is the displacement (distance between conjugate positions).
(7) Linear magnification, m = size of image/size of object = v/u [Note that by ‘size’ we mean linear size].
(8) Areal magnification = Area of image/Area of object = m2
(9) Condition for achromatic doublet (achromatic combination of two lenses):
ω1/f1 + ω2/f2 = 0 where ω1 and ω2 are the dispersive powers of the materials of the lenses. This can be written as ω1/f1 = -ω2/f2 indicating that one lens is converging while the other is diverging.
[Note that crown glass and flint glass lenses are used for achromatic doublets and if you require a converging doublet, the converging lens is to be made of crown glass since the dispersive power of crown glass is less than that of flint glass. You will then have smaller focal length for the converging lens to make the achromatic combination converging].
Let us now consider the following multiple choice question on refraction at a spherical surface:
A thick plano convex lens made of crown glass (refractive index 1.5) has a thickness of 3cm at its centre. The radius of curvature of its curved face is 5cm. An ink mark made at the centre of its plane face, when viewed normally through the curved face, appears to be at a distance ‘x’ from the curved face. Then, x is equal to
(a) 2cm (b) 2.1cm (c) 2.3cm
(d) 2.5cm (e) 2.7cm
The ray of light from the object O
(ink mark) gets refracted at the
interface between lens and air and
therefore appears to start from the
point I (Fig). So, I is the refracted
image of the object O. The object distance 'u' is PO and the image distance ‘v’ is PI. [P is the pole of the spherical surface].
We have, n2 /v – n1/u = (n2 – n1)/R so that 1/v – 1.5/(-3) = (1- 1.5)/(-5).
Note that we did not bother about the sign of the unknown quantity ‘v’. [In this problem we could have put negative sign for ‘v’ since the image I is on the same side as the object. Since we don’t apply sign to the unknown quantity ‘v’, we will obtain a negative value for ‘v’ on solving the problem]. The sign of ‘u’ is negative in accordance with the Cartesian convention. Since the incident ray is encountering a concave surface the radius of curvature is negative, in accordance with the convention.
Rearranging the above equation, we obtain 1/v = 0.5/5 – 1.5/3 = -6/15 from which v = -2.5cm.
Let us consider another MCQ:
An object is placed in front of a screen and a convex lens is placed at a position such that the size of the image formed is 9cm. When the lens is shifted through a distance of 20cm, the size of the image becomes 1cm. The focal length of the lens and the size of the object are respectively,
(a) 7.5cm and 3.5cm (b) 7.5cm and 4cm (c) 6cm and 3cm (d) 7.5cm and 3cm (e) 6cm and 2.5cm

If h1 and h2 are the sizes of the of the image in the two conjugate positions, the size of the object is given by h = √(h1h2) = √(9×1) = 3cm.
Considering the formation of the image in the first case, we have v/u = 9/3 so that v = 3u. Also, v = 20+u since v and u interchange in the conjugate positions. Therefore, 3u = 20+u from which u = 10cm. v = 20+u = 30cm.
Focal length f = uv/(u+v), since v is positive and u is negative in the equation, 1/f = 1/v - 1/u. Therefore, f = 10×30/(10+30) = 7.5cm.
[Note that the answer is positive as expected for a convex lens].
The correct option is (d).
The following simple question appeared in Kerala Medical Entrance 2001 test paper:
The focal length of an equi-convex lens in air is equal to either of its radii of curvature. The refractive index of the material of the lens is
(a) 4/3 (b) 2.5 (c) 0.8 (d) 1.25 (e) 1.5
You should note that the radius of curvature of an equi-convex (or equi-concave) lens is equal to its focal length if the refractive index is 1.5. You can check it by substituting R1 = R2 = f in the lens maker’s equation: 1/f = (n – 1) [1/f – 1/(-f)] = (n – 1)×2/f, from which n = 1.5. So the correct option is (e).
You should also note that the focal length of a plano-convex or plano-concave lens is twice the radius of its curved face if the refractive index is 1.5.
The following MCQ appeared in IIT 1999 question paper:
A concave lens of glass, refractive index 1.5, has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a
(a) convergent lens of focal length 3.5R (b) convergent lens of focal length 3.0R (c) divergent lens of focal length 3.5R (d) divergent lens of focal length 3.0R.
On substituting the given quantities in the lens maker’s equation, 1/f = (n2/n1 – 1)(1/R1 – 1/R2), we have 1/f = [(1.5/1.75)-1] [1/(-R) – 1/R)] = (-0.25/1.75) (-2/R) = 0.5/1.75R, from which f = 3.5R.
You should ensure that all your doubts on applying the sign convention are cleared before proceeding to the next question. In the lens maker’s equation, we applied negative sign to the radius of curvature (R) of the first surface since the concave lens presents a concave surface to the incident ray (and hence the centre of curvature is on the same side of the pole as the object is). The sign of the radius of curvature of the second surface is positive since it presents a convex surface to the ray proceeding towards it after the refraction at the first surface. We did not bother about the sign of the focal length of the lens, since it is the unknown quantity in this problem. The sign turns out to be positive, indicating that the concave lens behaves as a converging lens on immersing it in a denser liquid. The correct option therefore is (a).
You should note that a divergent lens becomes convergent (and a convergent lens becomes divergent) on immersing in a denser medium (of greater refractive index).
Let us consider a numerical example in this context:
A convergent lens made of crown glass (refractive index 1.5) has focal length 20cm in air. If it is immersed in a liquid of refractive index 1.60, its focal length will be
(a) 160cm (b) 100cm (c) -80cm (d) -100cm (e) -160cm
When the lens is in air, we have (from lens maker’s equation),
1/20 = [(1.5/1) – 1] (1/R1- 1/R2).
We are not bothered about the signs of R1 and R2 since they are unknown quantities.
Even though we know that the convergent lens will become divergent in the denser medium, we write lens maker’s equation without bothering about the sign of its unknown focal length in the liquid. If ‘f’ is the focal length in the liquid, we can write,
1/f = [(1.5/1.6) – 1] (1/R1 – 1/R2).
Dividing the first equation by the second, we obtain f/20 = 0.5×1.6/(-0.1) from which f = -160 cm [Option (e)].
Consider now the following MCQ:
The plane face of a plano-convex lens of focal length 20cm is silvered. The lens will then behave as a concave mirror of focal length
(a) 5cm (b) 10 cm (c) 20cm (d) 40cm (e) 80cm
When you direct a ray of light (through the un-silvered surface) in to this lens, the ray will be reflected by the silvered face. Therefore, the ray undergoes refraction twice at the lens and the effect is to make the lens a biconvex one, as though we are keeping two identical plano-convex lenses in contact, the only difference being that the ray is sent back. The focal length of the silvered lens therefore becomes half that of the un-silvered one. So, the answer is 10cm [Option (b)].
In this problem, the silvered surface, being plane, acts just to return the ray. If the curved surface were silvered, it would have acted as a spherical mirror complicating the problem. But, there is no significant complication and you can tackle the problem using the general expression for the resultant focal length (F) given by 1/F = 1/f + 1/fm + 1/f.
Or, 1/F= 2/f + 1/fm. Here ‘f’ is the focal length of the un-silvered lens while fm is the focal length of the mirror.
The power of the un-silvered lens appears twice and the power of the mirror appears once in this equation.
When you use this equation for solving the above question, the mirror term will vanish since the focal length of a plane mirror is infinite and you get F = f/2.
The following simple question appeared in Kerala Medical Entrance 2000 test paper:
A convex lens of focal length 40cm is in contact with a concave lens of focal length 25cm. The power of the combination in dioptre is
(a) -6.5 (b) -1.5 (c) +1.5 (d) +6.5 (e) +6.67
As you know, the power (expressed in dioptre) is the reciprocal of the focal length in metre. A convex lens (or, generally a converging lens) has positive power while a concave lens (or, generally a diverging lens ) has negative power. When you combine lenses in contact, the net power (P) of the combination is equal to the algebraic sum of the individual powers: P = P1+P2+P3+….
In the present case, P = (1/0.4) + (-1/0.25) = 2.5 – 4 = -1.5 dioptre.
The following MCQ appeared in the IIT Screening 2005 question paper:
Concave and convex lenses are placed touching each other. The ratio of magnitudes of their powers is 2:3. The focal length of the system is 30cm. Then the focal lengths of individual lenses are
(a) -75cm, 50cm (b) -15cm, 10cm (c) 75cm, 50cm (d) 75cm, -50cm
This too is a simple question. The power of the combination is 1/0.3 dioptre. [Note that the power of the combination is positive since the focal length of the combination is positive]. The concave lens has therefore smaller power and since the ratio of magnitudes of powers is 2:3, we can write -2k+3k = 1/0.3 where ‘k’ is the constant of proportionality. From this, k = 1/0.3 so that the powers of the concave and convex lenses are respectively – (2/0.3) and +(3/0.3). Therefore, their focal lengths are – (0.3/2)m and +(0.3/3)m. The focal lengths in cm are -15 and +10 [Option (b)].

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