The sign convention adopted widely in Optics is the Cartesian 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.

Occasionally, you may be given a ray diagram in which the incident ray may be proceeding from right to left. 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: pole to right positive and pole to left negative. Distances measured (from pole) upwards are positive and those measured downwards are negative but you will mostly encounter problems with leftward and rightward measurements.

While solving problems, you should apply the signs to all known quantities. The unknown quantities are left as they are in the formulae. You will be able to arrive at conclusions by interpreting the sign of the unknown quantity you finally arrive at 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.

Let us consider the following M.C.Q.:

1/20 = [1.5-1] [(1/R

1/f = [1.125 – 1] [(1/R

Note that we have ignored the signs of the unknown quantities f, R

The result shows that within water, the lens is still a converging lens. If the liquid has refractive index more than that of the lens, the lens will become diverging within the liquid. Let us consider such a case:

1/f = [0.9375 – 1] [(1/R

Dividing the first equation by the second, we obtain f = -160cm. As we get a negative value for the focal length, the lens behaves as a diverging lens within the given liquid [option (d)].

Let us discuss one more question:

1/F = 1/f + 1/fm + 1/f where ‘f’ is the focal length of the un-silvered lens and fm is the focal length of the silvered surface which makes a mirror. In the present problem since the silvered surface is plane, fm is infinity so that the above equation becomes

1/F = 1/f + 0 + 1/f = 2/f. Therefore, F = f/2.

But we have for the original un-silvered lens, 1/f = (μ – 1) [(1/R

Since F = f/2, the system has a focal length of R/2(μ – 1).The radius of curvature of a concave mirror is twice its focal length. Therefore, the system behaves like a concave mirror of radius of curvature R/(μ – 1). [Option (c)]. You will find multiple choice questions (with solution) on refraction at plane surfaces

Occasionally, you may be given a ray diagram in which the incident ray may be proceeding from right to left. 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: pole to right positive and pole to left negative. Distances measured (from pole) upwards are positive and those measured downwards are negative but you will mostly encounter problems with leftward and rightward measurements.

While solving problems, you should apply the signs to all known quantities. The unknown quantities are left as they are in the formulae. You will be able to arrive at conclusions by interpreting the sign of the unknown quantity you finally arrive at 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.

Let us consider the following M.C.Q.:

**A convex lens made of crown glass of refractive index 1.5 has focal length 20 cm. What will be its focal length within water (refractive index 4/3)?**

(a) 10 cm (b) 20 cm (c) 40 cm (d) 60 cm (e) 80 cm

Writing lens maker’s equation 1/f = [(μ(a) 10 cm (b) 20 cm (c) 40 cm (d) 60 cm (e) 80 cm

_{2}/μ_{1}) -1] [(1/R_{1}) - (1/R_{2})] for the two cases, we have,1/20 = [1.5-1] [(1/R

_{1}) - (1/R_{2})] for the first case, where f = +20cm (for convex lens), μ_{2}= 1.5 and μ_{1}= 1 (for air) and1/f = [1.125 – 1] [(1/R

_{1})-(1/R_{2})] for the second case where μ_{2}= 1.5 and μ_{1}= 4/3 (for water).Note that we have ignored the signs of the unknown quantities f, R

_{1}and R_{2}. On dividing the first equation by the second, we obtain f = 80 cm[option (e)].The result shows that within water, the lens is still a converging lens. If the liquid has refractive index more than that of the lens, the lens will become diverging within the liquid. Let us consider such a case:

**A converging lens of focal length 20cm is made of glass of refractive index 1.5. How will this lens behave if it is immersed in a liquid of refractive index 1.6?**

(a) As a converging lens of focal length 80cm (b) As a diverging lens of focal length 80cm (c) As a diverging lens of focal length 120cm (d) As a diverging lens of focal length 160cm (e) As a converging lens of focal length 120cm

The correct option to this question is (d). You can check this by writing the lens maker’s equation for the two cases as in the previous question. The first equation is unchanged: 1/20 = [1.5-1] [(1/R(a) As a converging lens of focal length 80cm (b) As a diverging lens of focal length 80cm (c) As a diverging lens of focal length 120cm (d) As a diverging lens of focal length 160cm (e) As a converging lens of focal length 120cm

_{}) - (1/R_{2})]. The second equation is1/f = [0.9375 – 1] [(1/R

_{1}) - (1/R_{2})] since μ_{2}/μ_{1}) = 1.5/1.6 = 0.9375.Dividing the first equation by the second, we obtain f = -160cm. As we get a negative value for the focal length, the lens behaves as a diverging lens within the given liquid [option (d)].

Let us discuss one more question:

**The plane face of a planoconvex lens is silvered. If the refractive index of the material of the lens is μ, and the radius of curvature of its curved face is R, then the system behaves like a concave mirror of radius of curvature**

(a) Rμ (b) R/μ (c) R/ (μ-1) (d) R/2(μ-1) (e) R(μ-1)

When one surface of a lens is silvered, the rays of light entering through the un-silvered surface are refracted first, reflected by the silvered surface next and finally are once more refracted. The effective focal length (F) of the system is given by(a) Rμ (b) R/μ (c) R/ (μ-1) (d) R/2(μ-1) (e) R(μ-1)

1/F = 1/f + 1/fm + 1/f where ‘f’ is the focal length of the un-silvered lens and fm is the focal length of the silvered surface which makes a mirror. In the present problem since the silvered surface is plane, fm is infinity so that the above equation becomes

1/F = 1/f + 0 + 1/f = 2/f. Therefore, F = f/2.

But we have for the original un-silvered lens, 1/f = (μ – 1) [(1/R

_{1}) - (1/R_{2})] where R_{1}= R and R_{2}= infinity. Therefore, f = R/(μ – 1)Since F = f/2, the system has a focal length of R/2(μ – 1).The radius of curvature of a concave mirror is twice its focal length. Therefore, the system behaves like a concave mirror of radius of curvature R/(μ – 1). [Option (c)]. You will find multiple choice questions (with solution) on refraction at plane surfaces

**here**as well as**here**
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