Today we will discuss two questions (MCQ) involving lenses. The first one appeared in Karnataka CET 2011 question paper and the second one appeared in Karnataka CET 2010 question paper.

(1) Two thin lenses have a combined power of + 9 D. When they are separated by a distance of 20 cm, their equivalent power becomes + 27/5 D. Their individual powers (in dioptre) are ……..

(1) 4, 5

(2) 3, 6

(3) 2, 7

(4) 1, 8

We have *P*_{1} + *P*_{2} = 9 where *P*_{1} and *P*_{2} are the powers of the two lenses.

[Note that combined power means 1/*f*_{1} + 1/*f*_{ 2 }which is the effective power when the lenses are in contact. Power is the reciprocal of focal length and when the lenses are in contact, we have 1/*F* = 1/*f*_{1} + 1/*f*_{ 2} where * f*_{1} and *f*_{2 }are the focal lengths of the lenses and *F * is the combined focal length].

When the lenses are separate by a distance *d*, the power becomes *P*_{1} + *P*_{2 }–* d** P*_{1}*P*_{2}

[This follows from the expression, 1/*F* = 1/*f*_{1} + 1/*f*_{ 2 }–* d**/f*_{1}*f*_{2}].

Therefore, we have

*P*_{1} + *P*_{2 }–* d** P*_{1}*P*_{2} = 27/5

Or, 9 –* *(0.2×* P*_{1}*P*_{2})_{ }= 27/5

[You have to substitute the separation in *metre* since the power is the reciprocal of *focal length in metre*].

From the above equation, 18/5 = 0.2×* P*_{1}*P*_{2}

Or, *P*_{1}*P*_{2} = 18

Evidently the values given in option (2) satisfy the above condition so that we have *P*_{1} = 3 and *P*_{2} = 6

(2) The focal length of a plano-convex lens is ‘*f*’ and its refractive index is 1.5. It is kept over a glass plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result, the effective focal length of the combination becomes 2*f*. Then the refractive index of the liquid is ……

(1) 1.5

(2) 2

(3) 1.25

(4) 1.33

In the case of a lens placed in air or vacuum, lens maker’s equation is

1*/f* = (*n* – 1)(1/*R*_{1} – 1/*R*_{2}) where n is the refractive index of the lens.

[General form of lens maker’s equation is 1*/f* = (*n*_{2}/*n*_{1} – 1)(1/*R*_{1} – 1/*R*_{2}) where ‘*f*’ is the focal length of the lens, *R*_{1} and *R*_{2} are the radii of curvature of its faces, *n*_{2} is its refractive index and *n*_{1} is the refractive index of the medium in which the lens is placed].

In the case of the plano-convex lens, the equation becomes

1*/f* = (*n* – 1)(1/*R*_{1 }– 1/∞) = (*n* – 1)(1/*R*_{1})

Since *n* = 1.5, *R*_{1} = 0.5*f* ……..(i)

The focal length of the combination (*F*) of the plano convex lens (of focal length *f*) and the *plano-concave* liquid lens (of focal length *f*_{ℓ})is given by

1/*F *= 1*/f + *1*/f*_{ℓ}

Therefore, 1/2*f* = 1*/f + *1*/f*_{ℓ}

This gives *f*_{ℓ} = –2*f*………(ii)

But from lens maker’s equation, the focal length of the plano-concave liquid lens is related to the refractive index (*n*_{ℓ}) of the material (liquid) as

1/*f*_{ℓ} = (*n*_{ℓ} – 1)(–1/*R*_{1 }– 1/∞) = (*n*_{ℓ} – 1)(– 1/*R*_{1})

[Note that the radius of the *concave* surface, which the liquid lens presents to the incident light, has to be *negative* in accordance with the Cartesian sign convention. Click on the label ‘Cartesian sign convention’ below this post to learn more about this convention].

Therefore, 1/(–2*f*)* =* – (*n*_{ℓ} – 1)/0.5*f** *, on substituting for *R*_{1} and *f*_{ℓ} from equations (i) and (ii).

This gives 2 *n*_{ℓ} = 2.5 so that *n*_{ℓ} = 1.25

## No comments:

## Post a Comment