Optics Concept Page - 18

Definition

Displacement method to find focal length of a lens

If the distance D of an object between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens between the object and the screen at which a sharp image of the object is formed on the screen. This method is called displacement method.
An illuminated object is set up in front of a lens and a focused image is formed on a screen.For a given separation of the object and screen it will be found that there are two positions where a clearly focused image can be formed. By the principle or reversibility these must be symmetrical between 0 and I. 
Using the notation shown: d=u+v$d=u+v$ and a=v−u$a=v-u$
Therefore: u=d−a2$u={\displaystyle \frac{d-a}{2}}$ and v=d+a2$v={\displaystyle \frac{d+a}{2}}$
Substituting in the lens equation gives:
2d−a+2d+a=1f${\displaystyle \frac{2}{d-a}}+{\displaystyle \frac{2}{d+a}}={\displaystyle \frac{1}{f}}$ and hence f=d2−a24d$f={\displaystyle \frac{d^2-a^2}{4d}}$

Example

Lens Maker's Formula

Consider a convex lens (or concave lens) of absolute refractive index μ2${\mu }_2$ to be placed in a rarer medium of absolute refractive index μ1${\mu }_1$. Considering the refraction of a point object on the surface XP1Y$X{P}_{1}Y$, the image is formed at I1$I_1$ who is at a distance of V1$V_1$.

CI1=P1I1=V1$C{I}_1=P_1{I}_1=V_1$ (as the lens is thin)

CC1=P1C1=R1$C{C}_1=P_1{C}_1=R_1$

CO=P1O=u$CO=P_{1}O=u$

It follows from the refraction due to convex spherical surface XP1Y$X{P}_{1}Y$

μ1−u+μ2v1=μ2−μ1R1..........(i)${\displaystyle \frac{{\mu }_1}{-u}}+{\displaystyle \frac{{\mu }_2}{v_1}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R_1}}..........(i)$

The refracted ray from A suffers a second refraction on the surface XP2Y$X{P}_{2}Y$ and emerges along BI. Therefore I is the final real image of O.

Here the object distance is

u=CI1≈P2I1=V$u=C{I}_1\approx P_2{I}_1=V$

P1P2$P_1{P}_2$ is very small

Let CI≈P2I=V$CI\approx P_{2}I=V$

(Final image distance)

Let R2$R_2$ be radius of curvature of second surface of the lens. It follows from refraction due to concave spherical surface from denser to rarer medium that

−μ2v1+μ1v=μ1−μ2R2=μ2−μ1−R2............(ii)${\displaystyle \frac{-{\mu }_2}{v_1}}+{\displaystyle \frac{{\mu }_1}{v}}={\displaystyle \frac{{\mu }_1-{\mu }_2}{R_2}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{-R_2}}............(ii)$

Adding (i)and(ii)$(i)and(ii)$

−μ1u+μ1v=(μ2−μ1)(1R1−1R2)${\displaystyle \frac{-{\mu }_1}{u}}+{\displaystyle \frac{{\mu }_1}{v}}=({\mu }_2-{\mu }_1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

or

μ1(1v−1u)=(μ2−μ1)(1R1−1R2)${\mu }_1({\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}})=({\mu }_2-{\mu }_1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

But 1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$ and μ2μ1=μ${\displaystyle \frac{{\mu }_2}{{\mu }_1}}=\mu$

Thus we get=

1f=(μ−1)(1R1−1R2)${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

Formula

Lens formula

Lens formula is an expression relating the image distance (v$v$), object distance (u$u$) and the focal length (f$f$) of a lens.
1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
Note:

• Utmost care should be taken regarding sign convention while using lens formula.
• Focal length of convex lens is taken as positive and concave lens as negative.

Definition

Lateral magnification for refraction at spherical surfaces

Assumptions:
θ1<<1${\theta }_1<<1$            θ2<<1${\theta }_2<<1$ 
n1θ1≃n2θ2$n_1{\theta }_1\simeq n_2{\theta }_2$
Using small angle approximation and Snell's law,
θ1≃h0d0${\theta }_1\simeq {\displaystyle \frac{h_0}{d_0}}$
θ2≃−hidi${\theta }_2\simeq {\displaystyle \frac{-h_i}{d_i}}$
n1h0d0≃−n2hidi${\displaystyle \frac{n_1{h}_0}{d_0}}\simeq {\displaystyle \frac{-n_2{h}_i}{d_i}}$
∴m=hih0=−n1din2d0$\therefore m={\displaystyle \frac{h_i}{h_0}}={\displaystyle \frac{-n_1{d}_i}{n_2{d}_0}}$

Definition

Dependence of focal length of a lens on the wavelength of incident light

High frequency waves, like blue colour, travel the slowest in any given medium, compared to the low frequency waves like red colour. (except for in vacuum, in which they all travel with the same speed). Thus, high frequency waves like blue colour, bend more (more refraction or more refractive index) compared to low frequency waves like red. Thus when incident on a convex (or concave) lens, blue colour would bend more and thus converge (or diverge) closer to the lens. Hence, blue colour would have a shorter focal length, compared to red colour.

Example

Focal Length of a combination of lenses immersed in a medium

Example: The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive index 1.5$1.5$ and has a focal length 10 cm$10cm$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. Then they are put in contact with their convex surfaces touching each other.  If this combination lens is immersed in water (refractive index=43$={\displaystyle \frac{4}{3}}$), then find its focal length (in cm$cm$).                       
Solution:
μ2=1.5${\mu }_2=1.5$ and fair=10 cm$f_{air}=10cm$
f$f$ of 2 plano convex lenses =f$=f$ of a biconvex lens =2fair=20 cm$=2f_{air}=20cm$
So, fairfliq${\displaystyle \frac{f_{air}}{f_{liq}}}$ = μgμl−1μg−1${\displaystyle \frac{{\displaystyle \frac{{\mu }_g}{{\mu }_l}}-1}{{\mu }_g-1}}$ = 1.125−11.5−1${\displaystyle \frac{1.125-1}{1.5-1}}$
And, fliq=80 cm$f_{liq}=80cm$
feq=f1f2f1+f2=80×80100=40 cm$f_{eq}={\displaystyle \frac{f_1{f}_2}{f_1+f_2}}={\displaystyle \frac{80\times 80}{100}}=40cm$

Diagram

Object distance vs image distance for spherical lens

For a convex lens

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

u−vvu=1f${\displaystyle \frac{u-v}{vu}}=\frac{1}{f}$

(vu)(u−v)=f=constant${\displaystyle \frac{(vu)}{(u-v)}}=f=constant$

The above equation looks like a hyperbola.

Example

Use relationship between two image lengths and object length for a thin lens when object and screen are fixed and lens is moved

Example:
A convex lens forms a real image 4cm$4cm$ long on a screen. When the lens is shifted to a new position without disturbing the object or the screen, again real image is formed on the screen which is 16cm$16cm$ long. Find the length of the object.

Solution:
If the image size be I1 and I2$I_{1}and{I}_2$,
The object size is given by, using displacement method.
OS=I1I2−−−−√$OS=\sqrt{I_1{I}_2}$
Thus, OS=64−−√$OS=\sqrt{64}$
Or, OS=8

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