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Q.Two transparent media of refractive indices n1n_1 and n2n_2 are separated by a spherical transparent surface. The rays of light incident on the surface get refracted into the medium on the other side. The laws of refraction are valid at each point of the spherical surface. A lens is a transparent optical medium bounded by two surfaces, at least one of which should be spherical. The focal length of a lens is determined by the radii of curvature (R1R_1 and R2R_2) of its two surfaces and the refractive index (nn) of the medium of the lens with respect to the surrounding medium. Depending on R1R_1 and R2R_2, a lens behaves as a diverging or a converging lens. The ability of a lens to diverge or converge a beam of light incident on it defines its power.

(a) An object is placed at the point B as shown in the figure. The object distance (uu) and the image distance (vv) are related as
(i) 1v−1u=(n2−n1n1)1R\dfrac{1}{v}-\dfrac{1}{u}=\left(\dfrac{n_2-n_1}{n_1}\right)\dfrac{1}{R}
(ii) 1v−1u=(n1−n2n2)1R\dfrac{1}{v}-\dfrac{1}{u}=\left(\dfrac{n_1-n_2}{n_2}\right)\dfrac{1}{R}
(iii) n2v−n1u=(n2−n1)R\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{(n_2-n_1)}{R}
(iv) n1v−n2u=(n1−n2)R\dfrac{n_1}{v}-\dfrac{n_2}{u}=\dfrac{(n_1-n_2)}{R}
(b) A point object is placed in air at a distance 'R' in front of a convex spherical refracting surface of radius of curvature R. If the medium on the other side of the surface is glass, then the image is :
(i) real and formed in glass.
(ii) real and formed in air.
(iii) virtual and formed in glass.
(iv) virtual and formed in air.
(c) An object is kept at 2F in front of an equiconvex lens. The image formed is :
(i) real and of the size of the object.
(ii) virtual and of the size of the object.
(iii) real and enlarged.
(iv) virtual and diminished.
(d) A thin converging lens of focal length 10 cm and a thin diverging lens of focal length 20 cm are placed coaxially in contact. The power of the combination is :
(i) −5-5 D
(ii) +5+5 D
(iii) +15+15 D
(iv) −15-15 D
(e) An equiconcave lens of focal length 'f' is cut into two identical parts along the dotted line as shown in the figure. The focal length of each part will be :
(i) f4\dfrac{f}{4}
(ii) f2\dfrac{f}{2}
(iii) ff
(iv) 2f2f
Figure — CBSE 2022 55/4/1 Q12
Figure
Figure — CBSE 2022 55/4/1 Q12
Figure
CBSECBSE Class XII Board 2022Subjective· 5mImportance★★★★★
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(a) The refraction formula at a single spherical surface depends on which medium the object and image lie in; for object in n2n_2 and image in n1n_1, n1v−n2u=n1−n2R\frac{n_1}{v} - \frac{n_2}{u} = \frac{n_1 - n_2}{R}. (b) Virtual image in air. (c) Real, same size. (d) +5+5 D. (e) Focal length remains ff.


(a) Refraction at a Single Spherical Surface

The key is to recognize which medium contains the object and which contains the image. The general refraction formula at a spherical surface separating two media is derived from Snell's law applied at every point on the surface, combined with the paraxial approximation (small angles).

The standard form when light travels from medium of refractive index n1n_1 (object side) to medium of refractive index n2n_2 (image side) is:

n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

where uu is the object distance (measured from the surface), vv is the image distance, and RR is the radius of curvature (positive if the center of curvature is on the image side).

Looking at the figure description: the object is at point B on the right (in medium n2n_2), and the spherical surface is concave to the left with center C on the left (in medium n1n_1). Light travels from n2n_2 (where the object is) into n1n_1 (where the image will form).

Figure — CBSE 2022 55/4/1 Q12
Figure — CBSE 2022 55/4/1 Q12

When the object is in medium n2n_2 and the image forms in medium n1n_1, we swap the roles:

n1v−n2u=n1−n2R\frac{n_1}{v} - \frac{n_2}{u} = \frac{n_1 - n_2}{R}

This matches option (iv).

Watch out

The most common mistake is blindly applying the formula without checking which medium contains the object. Always identify the direction of light propagation first.

The correct answer for (a) is (iv).


(b) Point Object at Distance RR from Convex Surface

A point object is placed in air (n1=1n_1 = 1) at distance u=Ru = R from a convex spherical surface of radius RR, with glass (n2>1n_2 > 1, typically ≈1.5\approx 1.5) on the other side.

Using the refraction formula (light going from air into glass):

n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

Substituting n1=1n_1 = 1, u=−Ru = -R (object distance is negative in the sign convention where distances are measured from the surface, with the object on the left):

n2v−1−R=n2−1R\frac{n_2}{v} - \frac{1}{-R} = \frac{n_2 - 1}{R}

n2v+1R=n2−1R\frac{n_2}{v} + \frac{1}{R} = \frac{n_2 - 1}{R}

n2v=n2−1−1R=n2−2R\frac{n_2}{v} = \frac{n_2 - 1 - 1}{R} = \frac{n_2 - 2}{R}

v=n2Rn2−2v = \frac{n_2 R}{n_2 - 2}

For typical glass with n2=1.5n_2 = 1.5:

v=1.5R1.5−2=1.5R−0.5=−3Rv = \frac{1.5R}{1.5 - 2} = \frac{1.5R}{-0.5} = -3R

The negative sign indicates the image is on the same side as the object (in air), making it virtual.

Tip

When n2<2n_2 < 2, the denominator n2−2n_2 - 2 is negative, guaranteeing a virtual image in air for this configuration.

The correct answer for (b) is (iv): virtual and formed in air.


(c) Object at 2F2F in Front of Equiconvex Lens

An equiconvex lens has equal radii of curvature on both surfaces. When an object is placed at distance u=2fu = 2f (twice the focal length), we use the thin lens equation:

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Substituting u=−2fu = -2f (negative in the standard sign convention):

1v−1−2f=1f\frac{1}{v} - \frac{1}{-2f} = \frac{1}{f}

1v+12f=1f\frac{1}{v} + \frac{1}{2f} = \frac{1}{f}

1v=1f−12f=12f\frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f}

v=2fv = 2f

The positive value means the image is real (on the opposite side of the lens). The magnification is:

m=vu=2f−2f=−1m = \frac{v}{u} = \frac{2f}{-2f} = -1

The magnitude ∣m∣=1|m| = 1 means the image is the same size as the object. The negative sign indicates it is inverted.

The correct answer for (c) is (i): real and of the size of the object.


(d) Power of Lens Combination

When two thin lenses are placed in contact, their powers add:

Ptotal=P1+P2P_{\text{total}} = P_1 + P_2

Power is the reciprocal of focal length in meters:

P=1f (in meters)P = \frac{1}{f \text{ (in meters)}}

For the converging lens with f1=10f_1 = 10 cm =0.10= 0.10 m:

P1=10.10=+10 DP_1 = \frac{1}{0.10} = +10 \text{ D}

For the diverging lens with f2=−20f_2 = -20 cm =−0.20= -0.20 m (negative for diverging): …

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