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NEET 2024 Answer Key With Solution for Code T1, T2, T3, T4, T5, T6

Lens Maker's Formula - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Lens Maker's formula is considered one the most difficult concept.

  • 29 Questions around this concept.

Solve by difficulty

A thin convex lens made from crown glass\left ( \mu =\frac{3}{2} \right ) has a focal length of f. When it is measured in two different liquids having refractive indices\frac{4}{3} and \frac{5}{3} it has the focal lengths f_{1} \: and\: f_{2}  respectively. The correct relation between the focal lengths is :

 

A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and a plane mirror are arranged as shown.  For an object kept at a distance of 60 cm from the convex lens, the final image, formed by the combination, is a real image, at a distance of :

To find the focal length of a convex mirror, a student records the following data :

Object Pin

Convex Lens Convex Mirror

Image Pin

22.2 cm 32.2 cm 45.8 cm 71.2 cm

The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small,  f1 and f2 are close to :

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following option best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is

A lens having focal length f and aperture of diameter d forms an image of intensity I.

Apreture of diameter \frac{d}{2}  in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

Diameter of a plano - convex lens is 6 cm and thickness at the centre is 3 mm. If speed of the light in material of lens is 2 x 108 m/s, the focal length of the lens is :

 

 

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An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1cm  thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object be shifted to be in sharp focus on film?

When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index:

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If a convex lens having focal length f and refractive index 1.5 is immersed in water then new focal length will be \left(\mu_{w}=4 / 3\right)

Concepts Covered - 1

Lens Maker's formula

Lens Maker's formula -

Derivation of Lens maker formula - 

Let us take a lens having refractive index = nand the surrounding is having refractive index = n1. Also let us assume that the lens is having two refracting surface having radii R1 and R2

                                                 

 

As we have learned the formula of refraction at a single spherical surface. Let us apply this on the surface (1), we get - 

                                                                          \frac{n_{2}}{v_{1}}-\frac{n_1}{u}=\frac{n_{2}-n_1}{R_{1}} \ldots(1)

Similarly for the second surface - 

                                                                          \frac{n_{1}}{v}-\frac{n_{2}}{v_{1}}=\frac{n_{1}-n_{2}}{R_{2}} \ldots(2)

Here, v1 is the position of image formed by the first surface and the same image will now act as object for the second surface.

Now adding equation (1) and (2),

\begin{array}{l}{\frac{n_{1}}{v}-\frac{n_{1}}{u}=\left(n_{2}-n_{1}\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]} \\ \\ {\Rightarrow \frac{1}{v}-\frac{1}{u}=\left(\frac{n_{2}}{n_{1}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]}\end{array}

Now we are going to arrange this equation in the desired for as -

                                                                           So, put , \ u = \infty \ and \ v = f

we get,

                                                                         \frac{1}{f}=\left(\frac{n_{2}}{n_1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]

 

                                                                          \mathbf{\frac{1}{f}=(\mu_{relative}-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)}

                                                         Where, 

                                                                                 \mu_{\mathrm{relative}}=\frac{n_{\mathrm{lens}}}{n_{\text {medium }}}

There are ceratin limitations of this lens maker’s formula - 

  • The lens should not be thick so that the space between the two refracting surfaces can be small.
  • The medium used on both sides of the lens should always be same.

 

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Lens Maker's formula

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