What is Lens Formula and Magnification? - Coding

Light is a kind of energy that can be seen with the naked eye. We observe objects and understand the world around us mostly via the use of light. Light travels in a straight path at an extremely fast speed of around 3 × 10⁸ ms. A small light source produces a strong shadow on an opaque object. This means that the light travels in a straight line and the route is referred to as a ray of light, and a grouping of rays is referred to as a beam of light.

A light ray indicates the direction of light propagation. When light strikes a surface between two transparent mediums, it reflects and refracts, causing light rays to bend. Light rays bend around the edge of obstruction as well, although the bending is relatively minimal due to the very short wavelength of light radiation. This is known as light diffraction.

Lens Maker Equation

The image distance can be computed using the lens formula and knowledge of the object distance and focal length. The Lens formula describes the relationship between the distance of an image I the distance of an object (o), and the focal length (f) of the lens in optics. The lens formula works for both convex and concave lenses.

$\begin{aligned}\dfrac{1}{v}-\dfrac{1}{u}&=\left(\dfrac{\mu_2}{\mu_1} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\end{aligned}$

The lens equation is used by lens makers employ to create lenses with desired focal lengths. Lenses with varying focal lengths are employed in a variety of optical devices. The focal length of a lens is determined by the radii of curvature of two surfaces and the refractive index of the lens material.

Consider a thin glass lens with a refractive index of μ₂ and curvature centres C₁ and C₂ with curvature radii R₁ and R₂. Let μ₁ represent the refractive index of the surrounding medium. When a point object ‘O’ is maintained on-axis at a distance u from the lens, the ray OP passes through the optical centre without deviation. If the other ray OA had not been refracted along with AB by the first surface, it would have arrived at the point I’. If the second surface didn`t exist. But, due to the second surface, the ray undergoes another refraction at point B and reaches point I.

Refraction through a convex lens

For the first refraction, the object distance is u, and the image distance is v’. By using the relation among u. v. μ₁, and R for refraction at a single curved (convex) surface.

$\dfrac{\mu_2}{v'}-\dfrac{\mu_1}{u}=\dfrac{(\mu_2-\mu_1)}{R_1}$ ……(1)

For the second refraction, I’ acts as a virtual object in lens medium of refractive index μ₂, to produce a final image at I in surrounding medium of refractive index μ_1.By using the relation among u. v. μ₁, and R for refraction at a single curved (convex) surface.

$\dfrac{\mu_1}{v} - \dfrac{\mu_2}{v'} = \dfrac{(\mu_1 - \mu_2)}{R_2}$ ……(2)

Adding the Equation (1) and (2), we get

$\begin{aligned}\dfrac{\mu_1}{v} - \dfrac{\mu_2}{u}&=\dfrac{\mu_2 - \mu_1}{R_1} + \dfrac{\mu_1 - \mu_2}{R_2}\\\dfrac{1}{v}-\dfrac{1}{u}&=\left(\dfrac{\mu_2 - \mu_1}{\mu_1}\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\\&=\left(\dfrac{\mu_2}{\mu_1} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)\end{aligned}$ ……(3)

Case 1: If the point object is at infinity, its image is formed at the focal point. i.e. if u = ∞, then v = f; therefore equation (3) becomes-

$\dfrac{1}{f} = \left(\dfrac{\mu_2}{\mu_1} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ ……(4)

This is the lens maker`s equation. The lens of any desired focal length can be produced by choosing proper values of R₁, R₂, μ₁, μ_2.

Case 2: When the lens is kept in an air medium, μ₁ = 1 and the equation (4) becomes –

$\dfrac{1}{f} = ({\mu_2} - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ ……(5)

Using the new Cartesian sign convention, we get a positive focal length for the convex lens and a negative focal length for the concave lens.

Comparing the equation (3) and (5), we get

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

This is the lens formula, which is used to find the focal length of a given lens practically.

Magnification

Magnification means making objects appear larger than they are. Following are the different cases to determine the magnification for different cases as:

Magnification produced by a lens (m)

The magnification (m) of a lens is defined as the ratio of the height of an image to the height of an object.

m = Height of the image / Height of the object

m = h_i / h_o

It is also presented in terms of image distance and object distance.

m = Image distance / Object Distance

m = v / u

It is equivalent to the image-to-object distance ratio.

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Magnification produced by a Convex Lens

We know the properties of convex lens that it is virtual and upright. Because a convex lens may create both virtual and actual pictures, the magnification produced by a convex lens can be either positive or negative. Magnification is beneficial for virtual images but detrimental for real images. i.e. Positive (+ve) for virtual image and Negative (-ve) for the real image.

Case 1: If the magnitude of magnification is less than one, it means the image is smaller than the object. |m|<1, the image is diminished.

Case 2: If the magnitude of magnification is greater than one, then the image is larger than the object.

|m|>1, the image is Magnified.

Case 3: If the magnitude of magnification is one, then the image is the same size as an object. |m| = 1, the image is same size as object.

Magnification by a concave lens

When the item is in front of the concave lens, the image is in front of the same object on the same side. The concave lens always produces a virtual, erect, and reduced image. Because concave lenses always generate virtual images, the magnification achieved by them is always positive and it always produces an image that is smaller than the object.

The magnitude of magnification is less than one, so the image is smaller than the object.

|m|<1, the image is diminished.

Below tabular representation indicates the magnification and nature of the image for different cases for different lenses as:

Lens

Properties

Object distance (u)

Image distance (v)

Magnitude in terms of ‘u’ and ‘v’

Magnitude in terms of image height and object height

Convex Lens

Image produced are virtual and

upright(real image)

-ve

Virtual Image: -ve

Real Image: +ve

Virtual Image: +ve

Real Image: -ve

|m|=1

|m|<1

|m|>1

when h_i=h_o

when h_i<h_o

when h_i>h_o

Concave

Image produced is always virtual

-ve

Virtual image: +ve

|m|<1

–

Sample Questions

Question 1: If a thin lens is dipped in water, will its focal length change?

Answer:

The focal length of a lens is determined by the refractive index of the lens’s material in relation to its surroundings, as determined by the lens maker’s formula.

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

When a thin lens is submerged in water, its relative refractive index decreases, and therefore its focal length increases (and the lens’s power decreases).

Let f_a be the focal length of a thin glass lens in air and f_w be the focal lengths of a thin glass lens in water.

Let _aμ_g and _wμ_g = _aμ_g / _aμ_wbe the relative refractive indices of glass with respect to air and water.

Then,

$\frac{1}{f_a} = (_a\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

$\frac{1}{f_w} = (_w\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

Dividing both equations we get,

$\frac{f_w}{f_a} = \frac{_a\mu_g - 1}{_w\mu_g - 1}$

Since, _aμ_g> _wμ_g , f_w > f_a

$f_w = \frac{_a\mu_g - 1}{_w\mu_g - 1} \times f_a$

Question 2: Define the focal length of a lens, give signs of the focal length of the concave and convex lens?

Answer:

The focal length of a lens is the distance between its optical center and its primary focus.

It is defined as the distance between the optical center and the second primary focus, so that the focal length of a convex lens is positive and that of a concave lens is negative.

Question 3: What is linear magnification produced by a lens? what is the significance of its sign?

Answer:

The linear magnification produced by a lens is the factor by which the image size is changed from the object size.

It is defined as the ratio of the height of an image to the height of an object.

m = Height of the image / Height of the object

m = hi / ho

The linear magnification indicator specifies whether the image is upright or inverted in relation to the object. The magnification of an upright picture is positive, whereas the magnification of an inverted image is negative.

Sample Problems

Problem 1: The radii of curvatures of a convex lens are 40cm and 50cm, calculate the focal length if the refractive index of its material is 2.1.

Solution:

Given that: Refractive index (μ) = 2.1

Radii R₁ = 40cm

Radii R₂ = -50cm (sign conventions)

$\frac{1}{f} = (\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

= (2.1 – 1) (1/40 – 1/(-50))

= 1.1 × 9 / 200

= 0.0495

1 / f = 1 / 0.0495

= 20.2020 cm

The focal length of convex lens is 20.2020 cm.

Problem 2: A convex lens forms a real and inverted image of an object 40cm from the lens. Where will be the object placed in front of the convex lens, if the image is of the same size as the object?

Solution:

Given that: Image distance (v) = 40cm

Lens type is convex and image is real image

Height of image is as same size as of object i.e., h_i = -h_o

Since, magnification (m) = h_i / h_o = v / u

= -h_o / h_o = v / u

v / u = -1

u = -1 × v

u = -40cm.

1/f = 1/v -1/u

= 1/40 – 1/(-40)

= 2/40

1/f = 1/20

f = 20cm = 0.2m

The object is placed 40 cm away from the convex lens.

Problem 3: A concave lens of the focal length of 20cm forms an image of a needle 15 cm from the lens. How far is the needle placed from the lens?

Solution:

Given that:

Focal length (f) = -20cm

Image distance (v) = -15 cm

1/f = 1/v – 1/u

or

1/u = 1/v – 1/f

= 1/(-15) – (1/-20)

= -20 + 15 / 300

= – (1/60)cm

u = -60 cm

The object is placed 60cm in front of the concave lens.

Problem 4: a concave lens made of glass has a focal length of 20cm in air. Find its focal length when immersed in water. Given that the refracting index of the glass lens is 1.5 and that of in water is 4.

Solution:

Let f_a be the focal length of lens in air

Let f_w be the focal length of lens in water.

Given that:

f_a = – 20cm (concave lens)

_aμ_g= 1.5

_aμ_w = 4

_wμ_g = _aμ_g / _aμ_w

= (3/2) / (4/1)

= 3/8

$\frac{1}{f_a} = (_a\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

$\frac{1}{f_w} = (_w\mu_g - 1)(\frac{1}{R_1}- \frac{1}{R_2})$

Dividing equation both the equations we get,

$\frac{f_w}{f_a} = \frac{_a\mu_g - 1}{_w\mu_g - 1}$

$f_w = \frac{_a\mu_g - 1}{_w\mu_g - 1} \times f_a$

= (1.5 – 1) / (0.375 – 1) × (-20)

= 0.5 / (-0.625) × (-20)

f_w = 16cm

Problem 5: An optical system uses two thin convex lenses in contact having an effective focal length of 30/4 cm. If one of the lenses has a focal length of 30cm, find the focal length of the other.

Solution:

Given that:

f = 30/4cm

f₁ = 30cm

1/f = 1/f₁ + 1/f₂

1/f₂ = 1/f – 1/f₁

= 1/(30/4) – 1/30

= 4/30 – 1/30

1/f₂ = 1/10

The focal length of the other lens, f₂ is equal to 10 cm.

Reffered: https://www.geeksforgeeks.org

Class 12

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