Consider the following MCQ on normal refraction:
Suppose a glass slab of thickness ‘t’ and refractive index ‘n’ is placed on a printed page of a book you are reading. The prints will
(a) appear to be shifted towards you by (t – t/n)
(b) appear to be shifted towards you by t/n
(c) appear to be shifted away from you by (t – t/n)
(d) appear to be shifted away from you by t/n
(e) not appear to be shifted
The prints will be shifted towards you since you are viewing the prints through a denser medium. The apparent thickness of the glass slab will be t/n where as its real thickness is t. The print in contact with the glass slab will appear to be raised (shifted towards you) by the difference (t – t/n) between the real thickness and the apparent thickness
You should note that an object placed at a distance equal to 2f from a converging lens or a concave mirror of focal length f will result in a real image at distance 2f. Consider the following MCQ which is meant to check your awareness of this as well as the apparent shift produced by a transparent slab:
A glowing light emitting diode (LED) is placed on the principal axis of a convex lens at a distance of 24 cm from it. The focal length of the lens is 10 cm. When a glass slab of thickness ‘t’ and refractive index 1.5 is interposed between the lens and the LED, a well defined real image of the LED is formed at 20 cm on the other side of the lens. What is the thickness of the glass slab?(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
(e) 16 cm
Since the real image is formed by the lens at a distance 2f (= 20cm), the object must appear for the lens to be located at a distance of 20 cm as shown in the figure in which O is the object, I is the image formed by normal refraction at the slab and I1 is the final real image produced by the lens. In other words, the glass slab must produce an apparent shift of (24–20) = 4 cm so that the image I formed by normal refraction serves as the object for the lens. [The glass slab can be placed anywhere between the lens and the LED to produce this shift]. Therefore we have
(t – t/n) = 4cm.
Substituting for n = 1.5, we have
(1.5t – t)/1.5 = 4 cm, from which t = 12 cm.
Light is incident normally on face BC of a glass prism ABC as shown. A liquid of refractive index ‘n’ is placed on the face AB of the prism. If the refractive index of glass is 3/2, the maximum value of n for which total internal reflection at the face AB will occur is
(a) √3
(b) √2
(c) 2/√3
(d) (3√3)/2
(e) (3√3)/4
The angle of incidence of the ray of light on the face AB is 60º (noting that the angle of incidence is the angle between the ray and the normal to AB at the point of incidence).When the refractive index of the liquid increases, the critical angle for total reflection at AB also increases in accordance with the equation
sin θc = 1/n' where n' is the refractive index of glass with respect to the liquid given by
n' = (Refractive index of glass)/ (Refractive index of liquid) = (3/2)/n = 3/2n.
Since the angle of incidence of 60º is to be the critical angle, the maximum possible value of n is given by
Sin 60º = 1/(3/2n) = 2n/3.
Or, √3/2 = 2n/3, from which n = (3√3)/4
You should note that a converging lens will become diverging (and a diverging lens will become converging) if the lens is immersed in a medium of refractive index greater than that of the material of the lens. Let us discuss a problem to high light this point:
A thin biconcave lens made of glass of refractive index 1.5 has focal length 20 cm in air. In a liquid of refractive index 1.6 it will behave as
(a) diverging lens of focal length 60 cm
(b) diverging lens of focal length 120 cm
(c) converging lens of focal length 60 cm
(d) converging lens of focal length 120 cm
(e) converging lens of focal length 160 cm
We have the lens maker’s equation,
1/f = [(n2/n1) – 1] (1/R1 – 1/R2) with usual notations.
A concave lens in air is diverging and f is negative, R1 is negative and R2 is positive in accordance with the Cartesian sign convention. Therefore we have
1/(–20) = [(1.5/1) – 1] (–1/R1 – 1/R2), when the lens is in air.
[We don’t know the values of R1 and R2. These are not required in this question. But you must note that in the case of equi-convex and equi-concave lenses of refractive index 1.5, the radius of curvature will be equal in value to the focal length. You can check this using lens maker’s equation].
When the lens is within the liquid, we have
1/f = [(1.5/1.6) – 1] (–1/R1 – 1/R2)
[We don’t know the sign of the focal length f. It will be obtained as positive or negative depnding on whether the lens is converging or diverging in the liquid].
From the above two equations (on dividing one by the other), we obtain
–f /20 = (0.5×1.6)/(–0.1), from which f = 160 cm.
The sign of ‘f’ is positive and hence the lens will behave as a converging lens of focal length 160 cm.
You will find some useful multiple choice questions at physicsplus
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