The lens Maker’s formula is usefull to design lenses

Chapter–9: Ray Optics and Optical Instruments

Case Study 1:

Read the following paragraph and answer the question

The lens Maker’s formula is usefull to design lenses of desired focal lengths using surfaces of suitable radii of curvature. The focal length also depends on the refractive index of the material of the lensand the surrounding medium. The refractive index depends on the wavelength of the light used. The power of the lens is related to the focal length

Answer the following questions based on the above

(i)  How will the power of a lens be affected with an increase of wavelength of light ?            (1)

(ii) The radius of curvature of two surfaces of a convex lens is R each. For what value of μ of its material, will its focal length become equal to R ?                     (1)

(iii) The focal length of a concave lens of μ = 1.5 is 20 cm in air. It is completely immersed in water of μ = 4/3. Calculate its focal length in water                          (2)


An object is placed in front of a lens which forms its erect image of magnification 3. The power of the lens is 5 D. Calculate the distance of the object and the image from the lens.                            [CBSE   2023]


(i) This implies that the power of lens decrease with the increase of wavelength.

(ii) Radius of curvature of two surfaces R for convex lens.

Refractive index of material μ = ?

\dfrac{1}{f} = (\dfrac{\mu_2}{\mu_1}-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})

The radius of curvature of two surfaces of convex lens is equal to its focal length.

(iii) Focal length of a concave lens, f = 20 cm, μ = 1.5

Refractive index of water \mu_w = \dfrac{4}{3}

Length of water, f_w = ?

\dfrac{1}{f} = (\dfrac{15}{133} - 1)(\dfrac{1}{20} - \dfrac{1}{-20})

\Rightarrow \dfrac{1}{f} = (\dfrac{0.17}{133})(\dfrac{1}{10})

\Rightarrow \dfrac{1}{f} = 0.127 \times \dfrac{1}{10}


(iiii) Given that, magnification (m) = 3

Power(P) = 5 D

Magnification, m = \dfrac{-v}{u} = 3

v = 3u

Then, \dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}

\Rightarrow \dfrac{1}{3u}-\dfrac{1}{u} = \dfrac{1}{f}

\Rightarrow \dfrac{1}{3 u} - \dfrac{1}{u} = 5

\Rightarrow \dfrac{-2}{3u} = 5

\Rightarrow u = \dfrac{-2}{3 \times 5} = \dfrac{-2}{15}.

and v = 3 \times \dfrac{-2}{15}

Hence, v = \dfrac{-2}{5}

Case Study 2:

Read the following paragaraph and answer the questions that follow

Types of Lenses and their combination

A convex or converging lens is thicker at the centre than at the edges. It converges a beam of light on refraction through it. It has a real focus Convex lens is of three types:  Double convex lens,Plane-convex lens and Concave-convex lens.

Concave lens is thinner at the centre  than at the edges. It diverges a beam of light on refraction through it. It has a virtual focus. Concave lenses are of three types: Double concave lens, Plane-concave lens and Convexo-concave lens.

When two thin lenses of focal length f_1 and f_2 are placed in contact with each other along their common principal axis, then the two lens system is regarded as a single lens of focal length f and \dfrac{1}{f} = \dfrac{1}{f_1}+\dfrac{1}{f_2}.

If several thin lenses of focal lengths f_1,f_2,f_3 . . . . f_n are placed in contact, then the effective focal length of the combination is given by

\dfrac{1}{f} = \dfrac{1}{f_1}+ \dfrac{1}{f_2} + . . . + \dfrac{1}{f_n}

And in terms of power, we can write

P = P_1 + P_2 + . . . + P_n

The value of focal length and power of lens must be used with proper sign consideration.

(i) Two thin lenses are kept  co-axially in contact with each other and the focal length of the combination is 80 cm. If the focal length of one lens is 20 cm, the focal length of the other would be               (1)

(a) – 26.7 cm           (b) 60 cm

(c)  80 cm                (d) 30 cm

(ii) A spherical air bubble is embedded in a piece of glass. For a ray of light passing through the bubble, it behaves like a                       (1)

(a) Converging lens     (b) Diverging lens

(c) Mirror                     (d) Thin plane sheet of glass

(iii) Lens generally used in magnifying glass is             (1)

(a) Single Concave lens

(b) Single convex lens

(c) Combination of convex lens of lower power and concave lens of lower focal length.

(d) Plano-concave lens

(IV) The magnification of an image by a convex lens is positive only, when the object is placed         (1)

(a) At its focus F

(b) Between F and 2F

(c) At 2F

(d) Between F and optical centre


A convex lens of 20 cm focal length forms a real image which is three times magnified. The distance of the object from the lens is              (1)

(a) 13.33 cm               (b) 14 cm

(c) 26.66 cm                (d) 25 cm

                    [CBSE SQP 2023]


Answer :-(i) (a)

Explanation:- Given, f = 80 cm, f_1 = 20 cm

For Combination of lens,

\dfrac{1}{f} = \dfrac{1}{f_1}+\dfrac{1}{f_2}.

\Rightarrow \dfrac{1}{80} = \dfrac{1}{20}+ \dfrac{1}{f_2}

\Rightarrow f_2 = - 26.7 cm

Answer:-(ii) (b)

Explanation:- A spherical bubble embedde in a piece of glass behaves as a diverging lens.

Answer:-(iii) (b)

 Explanation:-Single convex lens is used in magnifying glass.

Answer:- (iv) (d) 

Explanation:- When the object is placed between F and 2F infront of a convex lens, then magnification,

m = + ve



Explanation:- Given, f = +20 cm, m = -3 = v/u

⇒ v = -3u

By lens formula,

\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u}

\Rightarrow \dfrac{1}{-3u} - \dfrac{1}{u} = \dfrac{1}{20}

\Rightarrow u = - 26.66 cm

∴ Distance of the object = 26.66 cm

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