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Refraction by Lenses and Spherical Surfaces



Refraction Of Light - continued

Having studied refraction at plane surfaces and the concept of refractive index, we can now apply these principles to understand refraction at curved surfaces, particularly spherical lenses.


Refraction By Spherical Lenses

A lens is a transparent optical medium bounded by two surfaces, at least one of which is curved (usually spherical). Lenses work based on the phenomenon of refraction, bending light rays as they pass through. Lenses are essential components in many optical instruments.

Spherical lenses are of two main types:

These lenses are usually made of glass or plastic.


Terms Related to Spherical Lenses


Image Formation By Lenses

Similar to spherical mirrors, lenses form images whose characteristics (real/virtual, erect/inverted, magnified/diminished) depend on the type of lens and the object's position.


Image Formation In Lenses Using Ray Diagrams

Ray diagrams for lenses are used to trace the paths of light rays and locate the image. We use standard rays whose paths after refraction through a thin lens are known:

  1. A ray parallel to the principal axis, after refraction, passes through the second principal focus ($F_2$) of a convex lens, or appears to diverge from the second principal focus ($F_2$) of a concave lens.
  2. A ray passing through the first principal focus ($F_1$) of a convex lens, or directed towards the first principal focus ($F_1$) of a concave lens, after refraction, becomes parallel to the principal axis.
  3. A ray passing through the optical centre (O) of the lens goes through the lens undeviated.

To find the image, draw at least two of these rays from a point on the object. The point where the refracted rays intersect (for a real image) or appear to intersect (for a virtual image) is the corresponding point on the image.

Ray diagram for image formation by a convex lens.

(Image Placeholder: Ray diagram for a convex lens. Principal axis, Optical centre O, Foci F1, F2, and 2F1, 2F2 marked. An object is placed beyond 2F1. Ray 1 parallel to axis passes through F2. Ray 2 through O goes straight. The intersection of refracted rays between F2 and 2F2 forms a real, inverted, diminished image.)

Ray diagram for image formation by a concave lens.

(Image Placeholder: Ray diagram for a concave lens. Principal axis, Optical centre O, Foci F1, F2 marked (F2 on the same side as incident light for concave). An object is placed in front. Ray 1 parallel to axis appears to diverge from F2. Ray 3 through O goes straight. The intersection of the *extended* refracted ray 1 and ray 3 forms a virtual, erect, diminished image between O and F2 on the same side as the object.)


Sign Convention For Spherical Lenses

Consistent use of a sign convention is essential for applying lens formulas. The New Cartesian Sign Convention used for spherical mirrors is generally adapted for lenses with the optical centre as the origin.

  1. Origin: The optical centre (O) of the lens is taken as the origin (0,0).
  2. Principal Axis as X-axis: The principal axis is the X-axis.
  3. Distances Measured from Optical Centre: All distances are measured from the optical centre of the lens.
  4. Direction of Incident Light: Distances measured in the direction of the incident light are taken as positive.
  5. Opposite to Incident Light: Distances measured in the direction opposite to the direction of incident light are taken as negative.
  6. Heights Above Principal Axis: Heights measured upwards, perpendicular to the principal axis, are taken as positive.
  7. Heights Below Principal Axis: Heights measured downwards, perpendicular to the principal axis, are taken as negative.

Assuming incident light travels from left to right:


Lens Formula And Magnification

For thin spherical lenses, the relationship between object distance ($u$), image distance ($v$), and focal length ($f$) is given by the Lens Formula:

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

This formula is valid for both convex and concave lenses, for all positions of the object, provided the appropriate sign convention (New Cartesian Convention with origin at O, incident light from left) is used consistently.

The Magnification ($m$) produced by a spherical lens is defined as the ratio of the height of the image ($h_i$) to the height of the object ($h_o$).

$ m = \frac{\text{Height of Image}}{\text{Height of Object}} = \frac{h_i}{h_o} $

For thin spherical lenses, the magnification is also related to the object distance and image distance:

$ m = \frac{v}{u} $

The sign of magnification indicates the orientation of the image:

The magnitude of magnification indicates the relative size: $|m| > 1$ (magnified), $|m| < 1$ (diminished), $|m| = 1$ (same size).



Refraction At Spherical Surfaces And By Lenses

Understanding refraction at a single spherical surface is the stepping stone to deriving the lens maker's formula and the lens formula for thin lenses.


Refraction At A Spherical Surface

Consider a single spherical surface separating two transparent media with refractive indices $n_1$ and $n_2$. Let $R$ be the radius of curvature of the surface. Assume the surface is convex when viewed from medium 1 (where the object is placed). A point object O is placed in medium 1 at a distance $u$ from the pole P of the spherical surface. A ray from O, after refracting at the spherical surface, forms a point image I in medium 2 at a distance $v$ from P.

Diagram illustrating refraction at a convex spherical surface.

(Image Placeholder: A convex spherical surface separating Medium 1 (n1) and Medium 2 (n2). Pole P, Centre of Curvature C, Radius R. Object O in Medium 1, Image I in Medium 2. Show a ray from O incident at some point on the surface, refracts and goes to I. Show the normal to the surface passing through C.)

For paraxial rays (rays close to the principal axis and making small angles), the relationship between the object distance ($u$), image distance ($v$), radius of curvature ($R$), and refractive indices ($n_1$, $n_2$) is given by the formula for refraction at a single spherical surface:

$ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} $

This formula is valid using the New Cartesian Sign Convention with the pole P as the origin, distances to the right as positive, to the left as negative. Object in medium 1. $u$ is negative for real object on the left. $v$ is positive for real image on the right. $R$ is positive for a convex surface as its centre of curvature is on the right (if object is on the left). If the surface were concave from medium 1, $R$ would be negative.


Refraction By A Lens

A lens consists of two spherical (or one spherical and one plane) refracting surfaces. The image formed by the first surface acts as the object for the second surface, and the final image is formed by refraction at the second surface. Using the formula for refraction at a single spherical surface twice, once for each surface, and assuming the lens is thin (the thickness is negligible compared to the radii of curvature and object/image distances), we can derive the Lens Maker's Formula and the Lens Formula.


Lens Maker's Formula

The Lens Maker's Formula relates the focal length ($f$) of a thin lens to the refractive index of the lens material ($n_{lens}$), the refractive index of the surrounding medium ($n_{medium}$), and the radii of curvature of the two surfaces ($R_1$, $R_2$).

Let $n = n_{lens}/n_{medium}$ be the refractive index of the lens material relative to the surrounding medium. For a thin lens in air (where $n_{medium} \approx 1$, so $n \approx n_{lens}$):

$ \frac{1}{f} = (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

where $R_1$ is the radius of curvature of the first surface and $R_2$ is the radius of curvature of the second surface. The signs of $R_1$ and $R_2$ are determined using the sign convention: radius of a surface is positive if its centre of curvature is on the right side of the surface, and negative if its centre is on the left side, assuming light travels from left to right.

This formula is primarily used by lens manufacturers to design lenses of a desired focal length using a material of known refractive index.


Lens Formula

As discussed earlier, the Lens Formula relates $u, v,$ and $f$: $ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $ using the sign convention where $u$ is negative for real object, $v$ is positive for real image, and $f$ is positive for convex lens, negative for concave lens.


Power Of A Lens ($ P = 1/f $)

The power ($P$) of a lens is a measure of its ability to converge or diverge rays of light. It is defined as the reciprocal of its focal length (in metres).

$ P = \frac{1}{f} $

where $f$ is the focal length in metres. The SI unit of power of a lens is the dioptre (D). One dioptre is the power of a lens with a focal length of 1 metre.

$ 1 \text{ Dioptre (D)} = 1 \text{ m}^{-1} $

Using the sign convention for focal length:

Optometrists prescribe lenses for spectacles in dioptres (e.g., +2.5 D for reading glasses with convex lenses, -1.0 D for distant vision with concave lenses).

Example 1. A person needs a lens of power +1.5 D to correct their vision. What is the focal length and type of the lens?

Answer:

Power of the lens, $P = +1.5$ D.

The focal length $f$ is given by $P = 1/f$, where $f$ is in metres.

$ f = \frac{1}{P} = \frac{1}{+1.5 \text{ D}} = \frac{1}{1.5} $ metres.

$ f = \frac{1}{3/2} = \frac{2}{3} $ metres.

$ f = 0.667 $ metres, or 66.7 cm.

Since the power $P$ is positive, the focal length $f$ is positive. A lens with a positive focal length is a convex lens (converging lens).

The focal length of the lens is approximately 66.7 cm, and it is a convex lens.


Combination Of Thin Lenses In Contact ($ P_{eq} = P_1 + P_2 $)

When two or more thin lenses are placed in contact with each other coaxially, they form a combination that acts like a single lens. The power of this combination is simply the algebraic sum of the powers of the individual lenses.

Consider two thin lenses $L_1$ and $L_2$ with focal lengths $f_1$ and $f_2$, placed in contact. Let an object be placed at distance $u$. The first lens $L_1$ forms an image at $v_1$.

For lens $L_1$: $ \frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1} $

This image formed by $L_1$ acts as the object for the second lens $L_2$. Since the lenses are in contact, the distance of this object from $L_2$ is also $v_1$. If the image at $v_1$ is real, it acts as a real object for $L_2$ at distance $v_1$. If it is virtual, it acts as a virtual object at distance $v_1$. According to the sign convention, the object distance for the second lens is $u_2 = v_1$ (if image is on right of L1 and L2 is immediately after L1). However, the sign convention for the second lens requires distances to be measured from its optical center. Using the convention that positive distances are measured *in the direction of light travel AFTER* the lens, and considering the thin lens approximation, the object distance for the second lens is approximately $u_2 = v_1$. But careful sign convention handling is needed. Let's use the standard derivation where the image from L1 at $v_1$ becomes the object for L2, located at the same point but distances measured from L2. If the image from L1 is at distance $v_1$ to the right of L1, then for L2 (immediately after L1), the object distance is effectively $u_2 = v_1$. If this image is real (right of L1), light converges towards it. For L2, these converging rays are the *incident* rays. So, $u_2$ should be positive in the $1/v - 1/u$ formula? No, the incident rays for L2 come from the left, converging to a point on the right. A converging beam towards a point *after* the lens is a virtual object. So, if the image by L1 is at distance $v_1$ to the right, it's a virtual object for L2 at distance $v_1$, so $u_2 = v_1$. This gets complicated. Let's use a simplified approach consistent with the provided formula.

Assuming a thin lens combination, the effect of the combination is that of a single equivalent lens with focal length $f_{eq}$. An object at $u$ for the combination produces a final image at $v$.

For lens $L_1$, $1/v_1 - 1/u = 1/f_1$. Let the image from $L_1$ be at $I_1$. $I_1$ is the object for $L_2$. Distance of $I_1$ from $L_2$ is $v_1$ (if lenses are in contact). $L_2$ forms the final image at $v$. For lens $L_2$, $1/v - 1/v_1 = 1/f_2$ (the object distance for L2 is $v_1$ with reversed sign because it's on the right acting as object for L2). Adding the two equations: $ (1/v_1 - 1/u) + (1/v - 1/v_1) = 1/f_1 + 1/f_2 $ $ 1/v - 1/u = 1/f_1 + 1/f_2 $

Comparing this with the lens formula for the equivalent lens, $1/v - 1/u = 1/f_{eq}$:

$ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} $

This formula holds for two thin lenses in contact. If more lenses are in contact, the reciprocal of the equivalent focal length is the sum of the reciprocals of the individual focal lengths: $ \frac{1}{f_{eq}} = \sum_i \frac{1}{f_i} $.

Since the power of a lens is $P = 1/f$, the power of the combination of thin lenses in contact is the algebraic sum of the powers of the individual lenses:

$ P_{eq} = P_1 + P_2 + P_3 + ... = \sum_i P_i $

This formula is very useful for calculating the power or focal length of a system of lenses in contact. Remember to use the correct signs for the power of each lens (positive for convex, negative for concave).

The magnification produced by the combination of lenses in contact is the product of the magnifications produced by each lens: $m_{eq} = m_1 \times m_2 \times m_3 \times ...$

Example 2. A convex lens of focal length 30 cm is placed in contact with a concave lens of focal length 20 cm. Calculate the focal length and power of the combination.

Answer:

Focal length of convex lens, $f_1 = 30$ cm. Since it is convex, $f_1 = +30$ cm $= +0.30$ m.

Focal length of concave lens, $f_2 = 20$ cm. Since it is concave, $f_2 = -20$ cm $= -0.20$ m.

Power of individual lenses:

$ P_1 = \frac{1}{f_1} = \frac{1}{0.30 \text{ m}} = +3.33 $ D.

$ P_2 = \frac{1}{f_2} = \frac{1}{-0.20 \text{ m}} = -5.00 $ D.

Power of the combination ($P_{eq}$):

$ P_{eq} = P_1 + P_2 $

$ P_{eq} = +3.33 \text{ D} + (-5.00 \text{ D}) = -1.67 $ D.

Focal length of the combination ($f_{eq}$):

$ f_{eq} = \frac{1}{P_{eq}} = \frac{1}{-1.67 \text{ D}} $

$ f_{eq} \approx -0.5988 $ metres.

$ f_{eq} \approx -59.88 $ cm.

The focal length of the combination is approximately -59.9 cm, and the power is -1.67 D. Since the focal length is negative (or power is negative), the combination behaves as a concave lens (diverging lens).