Inference

The following is a comprehensive response to your queries, drawing exclusively from the provided source materials.

33. What is interference? What are the conditions for interference?

Interference is the modification of intensity obtained by the superposition of two or more waves of light. This phenomenon is the result of the interaction of light coming from two different wavefronts that originated from the same source.

Interference is governed by the general principle of superposition of waves. Because of this disturbance, the resultant amplitude and intensity may vary from the sum of those contributed by the individual beams acting simultaneously.

• If the resultant amplitude of the superposition is zero, the waves interfere destructively.
• If the resultant amplitude is greater than zero, they have constructive interference.

Conditions for Sustained Interference:

For producing a sustained and visible interference pattern, certain conditions must be met:

1. Coherent Sources: The two sources must be coherent, meaning they must have an identical frequency and a constant initial phase difference. Two independent sources cannot be used because they do not exist in nature.
2. Redistribution of Intensity: Interference itself is the redistribution of intensity in the region of superposition.

34. Derive the condition for maxima and minima due to interference of lights reflected from thin films of uniform thickness.

When light is reflected from a thin film of uniform thickness, the optical path difference ($\Delta$) between the two interfering rays (reflected from the top and bottom surfaces) is generally given by $2\mu t\cos r$.

However, when light reflects at the boundary of a rarer medium (Air) to a denser medium (Film), a phase change of $\pi$ occurs, which corresponds to an additional path difference of $\lambda /2$. This phase change occurs at the surface of the denser medium.

Thus, the total effective path difference in the reflected system is: $\Delta =2\mu t\cos r+\frac{\lambda }{2}\text{ or }2\mu t\cos r-\frac{\lambda }{2}$.

Where:

• $\mu$ is the refractive index of the film.
• $t$ is the thickness of the film.
• $r$ is the angle of refraction.

Conditions for Maxima (Constructive Interference/Bright Fringe):

The two rays will interfere constructively if the path difference is an integral multiple of the wavelength ($n\lambda$). $\Delta =n\lambda$ $2\mu t\cos r+\frac{\lambda }{2}=n\lambda \text{ or }2\mu t\cos r=(n-\frac{1}{2})\lambda$ Thus, the condition for maxima is: $2\mu \mathbf{t}\cos \mathbf{r}=(2n+1)\frac{\lambda }{2}$ where $n=0,1,2,3,4,...$.

Conditions for Minima (Destructive Interference/Dark Fringe):

The two rays will interfere destructively if the path difference is an odd multiple of half the wavelength ($(2n+1)\frac{\lambda }{2}$). $\Delta =(2n+1)\frac{\lambda }{2}$ $2\mu t\cos r+\frac{\lambda }{2}=(2n+1)\frac{\lambda }{2}\text{ or }2\mu t\cos r=n\lambda$ Thus, the condition for minima is: $2\mu \mathbf{t}\cos \mathbf{r}=\mathbf{n}\lambda$ where $n=0,1,2,3,...$.

35. Derive the condition for maxima and minima due to interference of lights transmitted from thin films of uniform thickness.

When light is transmitted through a thin film, the rays interfere upon exiting the film. The effective path difference ($\Delta$) between the two rays is calculated using geometry, but crucially, no phase change occurs when light is transmitted.

The effective path difference is simply: $\Delta =2\mu t\cos r$.

Conditions for Maxima (Constructive Interference/Bright Fringe):

The film appears bright in transmitted light when the path difference is an integral multiple of the wavelength ($n\lambda$). $\Delta =n\lambda$ Thus, the condition for maxima is: $2\mu \mathbf{t}\cos \mathbf{r}=\mathbf{n}\lambda$ where $n=0,1,2,3,...$.

Conditions for Minima (Destructive Interference/Dark Fringe):

The film appears dark in transmitted light when the path difference is an odd multiple of half the wavelength. $\Delta =(2n+1)\frac{\lambda }{2}$ Thus, the condition for minima is: $2\mu \mathbf{t}\cos \mathbf{r}=(2n+1)\frac{\lambda }{2}$ where $n=0,1,2,3,4,...$.

36. Why do we see colors in thin film when it is exposed to sunlight?

The phenomenon of colors seen in thin films (like soap bubbles or oil films on water) when exposed to sunlight originates from the interference of light.

1. Polychromatic Source: Sunlight (white light) consists of a continuous range of wavelengths ($\lambda$).
2. Path Difference Dependence: The path difference ($\Delta =2\mu t\cos r$) between the interfering rays (reflected from the top and bottom surfaces) depends on three factors: refractive index ($\mu$), thickness ($t$) of the film, and the inclination of the incident ray ($r$).
3. Selective Reflection: At a particular point on the film (fixed $t$) and for a specific observation direction ($r$), only certain wavelengths will satisfy the condition for maximum reflection (constructive interference).
   • Since only those specific wavelengths are reflected to the eye (and the rest are transmitted), the reflected light appears intensely colored.
4. Varying Conditions: Because the film thickness often varies, different wavelengths satisfy the interference condition at different points, leading to a visible display of multiple colors.

37. Why an extensive thin film appears black in reflected light?

An extensive thin film appears black in reflected light if its thickness ($t$) is excessively small compared to the wavelength ($\lambda$) of light (i.e., $2\mu t\ll \lambda$).

1. Negligible Thickness Term: When the film is extremely thin, the term $2\mu t\cos r$ can be neglected in calculating the path difference.
2. Effective Path Difference: In the reflected system, the only remaining effective path difference is that introduced by the phase change upon reflection, which is $\frac{\lambda }{2}$.
3. Destructive Condition: This path difference, $\frac{\lambda }{2}$, satisfies the condition for minimum intensity or destructive interference (which is $\Delta =(2n+1)\frac{\lambda }{2}$). When $n=0$, $\Delta =\frac{\lambda }{2}$.
4. Absence of Light: Since the condition for minimum intensity is satisfied, all wavelengths in the incident light are subject to destructive interference, meaning that every wavelength is absent in the reflected system, and the film therefore appears black in reflected light.

38. Why the fringes in the wedge shaped film are straight? State conditions for maxima and minima for interference in wedge shaped films.

Why the fringes are straight:

In a wedge-shaped film, the fringes are the locus of points that satisfy the condition for a maximum or a minimum. For interference to occur (either constructively or destructively) for a specific order ($n$) and a specific wavelength ($\lambda$), the path difference must remain constant.

The path difference depends on $\mu ,t,r,$ and $\theta$. Since the refractive index ($\mu$) and the wedge angle ($\theta$) are constant, and if we consider illumination by parallel monochromatic light at a constant angle of incidence ($r$ is constant), the path difference remains constant only if the thickness ($t$) of the film is constant.

The locus of points of constant film thickness in a wedge-shaped film forms straight lines parallel to the thin edge of the wedge. Therefore, the interference maxima and minima fringes are straight and parallel lines.

Conditions for Maxima and Minima (Reflected System):

The total path difference ($\Delta$) in a wedge-shaped film is given by: $\Delta =2\mu t\cos (r+\theta )+\frac{\lambda }{2}$.

1. Condition for Maxima (Constructive Interference): $2\mu \mathbf{t}\cos (\mathbf{r}+\theta )=(2n-1)\frac{\lambda }{2}$ where $n=1,2,3,4,...$.

2. Condition for Minima (Destructive Interference): $2\mu \mathbf{t}\cos (\mathbf{r}+\theta )=\mathbf{n}\lambda$ where $n=0,1,2,3,...$.

39. Whys are Newton’s ring are circular in nature and wedge shaped films are straight?

The difference in fringe shape stems directly from how the thickness of the enclosed air film varies geometrically in each setup.

Feature	Newton’s Rings	Wedge Shaped Films
Film Geometry	Air film is formed between a plano-convex lens and a plane glass plate. The thickness ($t$) of the film gradually increases with the distance from the point of contact.	Air film is formed between two plates inclined at a small angle ($\theta$). The thickness ($t$) increases uniformly starting from zero at the edge.
Fringe Shape	Circular.	Straight.
Reason	The locus of all points having constant film thickness ($t$) forms a circle centered at the point of contact. Since fringes are the locus of constant path difference (and thus constant $t$), the rings are circular.	The locus of all points having constant film thickness ($t$) forms a straight line parallel to the edge of the wedge. Hence, the fringes are straight and parallel.

40. Show in case on Newton’s ring the Diameter, $D\propto \sqrt{n}$ (dark ring), $D\propto \sqrt{(2n-1)}$ (bright ring)

The thickness ($t$) of the air film at a radius $r$ in Newton’s rings is related to the radius of curvature of the lens ($R$) by the approximate geometrical relation $r^2\approx 2Rt$. Since the diameter $D=2r$, we have $r=D/2$, leading to $t=\frac{D^2}{8R}$. Assuming the medium is air ($\mu =1$) and incidence is normal ($\cos r=1$).

1. Diameter of Dark Rings ($D_n\propto \sqrt{n}$):

The condition for a dark ring (minimum intensity) in the reflected system is: $2\mu t=n\lambda$ Substituting $t=\frac{D_n^2}{8R}$ and setting $\mu =1$: $2\left(\frac{D_n^2}{8R}\right)=n\lambda$ $\frac{D_n^2}{4R}=n\lambda$ $D_n^2=4nR\lambda$ Since $R$ and $\lambda$ are constant, we have: ${\mathbf{D}}_{\mathbf{n}}^2\propto \mathbf{n}\text{ or }{\mathbf{D}}_{\mathbf{n}}\propto \sqrt{\mathbf{n}}$ The diameter of the dark ring is proportional to the square root of the natural number $n$.

2. Diameter of Bright Rings ($D_n\propto \sqrt{2n-1}$):

The condition for a bright ring (maximum intensity) in the reflected system is: $2\mu t=(2n-1)\frac{\lambda }{2}$ Substituting $t=\frac{D_n^2}{8R}$ and setting $\mu =1$: $2\left(\frac{D_n^2}{8R}\right)=(2n-1)\frac{\lambda }{2}$ $\frac{D_n^2}{4R}=(2n-1)\frac{\lambda }{2}$ $D_n^2=2(2n-1)R\lambda$ Since $R$ and $\lambda$ are constant, we have: ${\mathbf{D}}_{\mathbf{n}}^2\propto (2n-1)\text{ or }{\mathbf{D}}_{\mathbf{n}}\propto \sqrt{2n-1}$ The diameter of the bright ring is proportional to the square root of the odd natural number $(2n-1)$.

41. Why Newton’s ring are circular and centre of interference pattern is dark?

Why Newton’s Rings are Circular:

Newton’s rings are formed by the air film enclosed between a plano-convex lens and a plane glass plate. The thickness of the film gradually increases as the distance from the point of contact increases.

For any fringe (maximum or minimum) to appear, the optical path difference ($2\mu t\cos r$) must be constant. Since the film medium ($\mu$) and the angle of incidence ($r$) are constant (due to normal incidence), the film thickness ($t$) must be constant for a given fringe.

The points having the same thickness ($t$) lie on a circle having the point of contact as its center. Therefore, the interference pattern consists of concentric dark and bright rings.

Why the Centre of the Interference Pattern is Dark (in reflected light):

The center of the Newton’s ring pattern corresponds to the exact point of contact between the lens and the glass plate.

1. Zero Thickness: At the point of contact, the thickness of the air film ($t$) is zero.
2. Phase Change: The interference occurs between rays reflected from the upper surface (air to lens/film, rarer to denser) and the lower surface (film to plate, denser to rarer for the air film case).
   • The reflection from the lower surface of the lens (air film boundary) occurs at a rarer medium boundary (assuming ${\mu }_{\text{air}}<{\mu }_{\text{lens}}$), leading to no phase change.
   • The reflection from the upper surface of the plate (air film boundary) occurs at a denser medium boundary (assuming ${\mu }_{\text{air}}<{\mu }_{\text{glass}}$), introducing a $\pi$ phase change, corresponding to an additional path difference of $\lambda /2$.
3. Path Difference at Center: For $t=0$ and normal incidence ($r=0$), the total path difference is: $\Delta =2\mu t\cos r+\frac{\lambda }{2}=0+\frac{\lambda }{2}=\frac{\lambda }{2}$.
4. Minimum Intensity: This path difference $\lambda /2$ satisfies the condition for minimum intensity (destructive interference, $2\mu t\cos r=n\lambda$ where $n=0$).
5. Result: Since the condition for minimum intensity is satisfied, the central spot is dark.

42. Explain why Newton’s ring are unequally spaced?

Newton’s rings are unequally spaced because the separation distance between consecutive dark (or bright) rings changes as the radius of the rings increases.

1. Relationship to Order: The diameters of the dark rings ($D_n$) are proportional to the square root of natural numbers ($D_n\propto \sqrt{n}$). $D_n^2=4nR\lambda$ The diameters of the bright rings are proportional to the square root of odd natural numbers ($D_n\propto \sqrt{2n-1}$).
2. Spacing Calculation: The spacing is determined by the difference between the diameters of consecutive rings.
   • For dark rings, the spacing increases as $\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4},\dots$ but the difference between successive square roots decreases as $n$ increases: $\text{Spacing}=D_{n+1}-D_n=\sqrt{4(n+1)R\lambda }-\sqrt{4nR\lambda }=\sqrt{4R\lambda }(\sqrt{n+1}-\sqrt{n})$
3. Unequal Separation: Since the quantity $(\sqrt{n+1}-\sqrt{n})$ decreases as $n$ increases, the rings become progressively closer together (unequally spaced) as the order of the ring increases.

43. What do we understand by anti-reflection coating? Derive the condition with proper diagram.

Anti-Reflection (AR) Coating:

An AR coating is a very thin, transparent film of proper thickness deposited on an optical surface (like a lens or telescope component) specifically to reduce the amount of light reflected from that surface. Uncoated optical surfaces typically reflect around 4% of light per surface, which is highly undesirable, especially in multi-element instruments.

Condition for Anti-Reflection Coating (Destructive Interference):

An AR coating works by utilizing destructive interference in the reflected light.

• Setup: Consider a ray incident normally on a film of refractive index $n_1$ coated on a glass plate of refractive index $n_2$ (See Figure 2.6.1 in the sources, which illustrates this setup). We require $n_{\text{air}}<n_1<n_2$ for optimal results.
• Interfering Rays: The incoming ray splits into two main reflected rays, $R_1$ (reflected from the air-film interface) and $R_2$ (reflected from the film-glass interface).
• Phase Change: If the refractive index condition ($n_{\text{air}}<n_1<n_2$) is met, both reflection events (Air $\to$ Film, Film $\to$ Glass) involve reflection from a boundary with a denser medium, meaning both rays $R_1$ and $R_2$ suffer a phase change of $\pi$. Therefore, there is zero phase difference ($\Delta \varphi =0$) due to reflection alone.
• Condition for Destructive Interference: To achieve minimum reflection (destructive interference), the path difference between $R_1$ and $R_2$ must be an odd multiple of $\lambda /2$. $2\mu t=(2n+1)\frac{\lambda }{2}$ Where $n=0,1,2,3,\dots$, and $\mu$ is the refractive index of the film ($n_1$) and $t$ is the thickness.

The simplest condition occurs when $n=0$, requiring the film thickness to be a quarter of the wavelength within the medium: $2\mu \mathbf{t}=\frac{\lambda }{2}$ $\mathbf{t}=\frac{\lambda }{4{\mu }_1}\text{ or }{\mu }_1\mathbf{t}=\frac{\lambda }{4}$.

This is the condition for the thickness of an anti-reflection coating.

(Note: The sources mention that ${\text{MgF}}_2$ with refractive index 1.38 is often used, as its R.I. is intermediate between air (1) and glass (1.5).)

44. What is highly reflected thin film? Why we need it?

Highly Reflected Thin Film:

A highly reflected thin film is a coating designed to maximize the intensity of reflected light by ensuring that the rays reflected from the film boundaries interfere constructively.

This effect is typically achieved when a thin film of thickness $t$ is applied such that the optical path difference leads to maximum reflection. For example, if the thickness $t$ is chosen such that $t=\lambda /(2\mu )$ or $t=3\lambda /(2\mu )$ (odd multiple of half wavelength optical thickness), this often results in maximum reflected intensity when considering the necessary phase changes. (Specifically, a thin film of thickness $t=\lambda /2$ creates constructive interference in reflection if the reflection criteria are met, resulting in a highly reflecting film).

Why we need it:

Highly reflecting films are necessary to create highly reflective mirrors, especially those used in optical systems where maximum reflectivity is critical.

A major application is in LASER technology. A laser cavity requires two mirrors surrounding the active medium. One mirror is typically highly reflective (sometimes 100%), ensuring that light waves move back and forth, building up the required intensity via constructive interference and stimulated emission.

This response explains the formation of Newton’s Rings and derives the formula used to determine the refractive index of a liquid based on the measurements of these rings.

Explanation of Newton’s Rings

Newton’s Rings are an interference pattern first studied by Newton. The rings are produced as a result of the interference between light waves reflected from the boundaries of a thin film of air (or liquid).

Experimental Setup and Formation:

1. Geometry: Newton’s rings are formed by placing a plano-convex lens (L) of large radius of curvature upon a plane glass plate (P). This arrangement encloses a thin air film whose thickness ($t$) is zero at the point of contact and gradually increases outwards along any radius.
2. Illumination: When monochromatic light is allowed to fall normally on this system, concentric dark and bright rings are formed, having the center dark.
3. Interference Source (Ray Diagram Description): The interference takes place between two sets of reflected rays:
   • Ray 1: Light reflected from the lower surface of the lens (the upper boundary of the air film). This reflection is from a rarer medium (Air) to a denser medium (Lens/Glass) and results in a phase change of $\pi$, equivalent to an extra path difference of $\lambda /2$.
   • Ray 2: Light reflected from the upper surface of the plane glass plate (the lower boundary of the air film). This reflection is from a denser medium (Air film) to a denser medium (Glass plate, if ${\mu }_{\text{glass}}>{\mu }_{\text{air}}$), causing no phase change.
4. Pattern: Since the rings are the locus of points of constant film thickness, and the thickness increases radially outward, the interference pattern consists of concentric circular rings.

(Note: The formation mechanism is illustrated in the sources by Fig. 1.9.1 and Fig. 1.9.2.)

Derivation of the Formula for Refractive Index ($\mu$) of a Liquid

The determination of the refractive index of a liquid involves comparing the diameters of the Newton’s rings when the space between the lens and plate is filled with air versus when it is filled with the liquid.

I. Case: Air Film (${\mu }_{air}=1$)

For an air film under normal incidence, the condition for the $n^{th}$ dark ring is: $2\mu t=n\lambda$ Since ${\mu }_{air}=1$ and $t=\frac{D_n^2}{8R}$ (where $R$ is the radius of curvature of the lens and $D_n$ is the diameter of the $n^{th}$ dark ring), the condition becomes: $\frac{D_{n,\text{air}}^2}{4R}=n\lambda$ $D_{n,\text{air}}^2=4nR\lambda$

For the $(n+p{)}^{th}$ dark ring in air, where $p$ is an integer representing the number of intervals counted beyond $n$: $D_{(n+p),\text{air}}^2=4(n+p)R\lambda$ Subtracting the equations: $D_{(n+p),\text{air}}^2-D_{n,\text{air}}^2=4(n+p)R\lambda -4nR\lambda$ $D_{(n+p),\text{air}}^2-D_{n,\text{air}}^2=4pR\lambda \dots (1)$

II. Case: Liquid Film (Refractive Index = $\mu$)

When a liquid of refractive index $\mu$ is introduced between the lens and the plate, assuming the liquid is rarer than the glass plate ($n_{\text{liquid}}<n_{\text{glass}}$) or if the experiment is set up such that the phase change occurs at the air/lens interface and not the liquid/glass interface, the fundamental condition for the $n^{th}$ dark ring remains: $2\mu t=n\lambda$ Substituting $t=\frac{D_{\text{n, liquid}}^2}{8R}$ (where $D_{\text{n, liquid}}$ is the diameter in liquid): $2\mu \left(\frac{D_{\text{n, liquid}}^2}{8R}\right)=n\lambda$ $D_{\text{n, liquid}}^2=\frac{4nR\lambda }{\mu }$

For the $(n+p{)}^{th}$ dark ring in the liquid: $D_{(n+p),\text{liquid}}^2=\frac{4(n+p)R\lambda }{\mu }$ Subtracting the equations for the $(n+p{)}^{th}$ and $n^{th}$ dark rings in liquid: $D_{(n+p),\text{liquid}}^2-D_{n,\text{liquid}}^2=\frac{4pR\lambda }{\mu }\dots (2)$

III. Determining Refractive Index

To find the refractive index $\mu$ of the liquid, we divide Equation (1) by Equation (2):

$\mu =\frac{D_{(n+p),\text{air}}^2-D_{n,\text{air}}^2}{D_{(n+p),\text{liquid}}^2-D_{n,\text{liquid}}^2}$

Thus, the refractive index $\mu$ is determined by the ratio of the difference between the squares of the diameters of two corresponding dark rings measured in air, to the difference between the squares of the diameters of the same dark rings measured in the liquid.

This response addresses your remaining queries based on the provided source materials and previous derivations.

46. With Newton’s ring experiment explain how to determine the refractive index of an unknown liquid.

The refractive index ($\mu$) of an unknown liquid is determined by comparing the diameter of Newton’s dark rings when the space between the plano-convex lens and the glass plate is filled with air (where ${\mu }_{\text{air}}\approx 1$) versus when it is filled with the test liquid.

Procedure for Determination:

1. Measurement in Air: The Newton’s ring experiment is performed with air filling the space. The diameters of a certain $n^{th}$ dark ring ($D_{n,\text{air}}$) and an $(n+p{)}^{th}$ dark ring ($D_{(n+p),\text{air}}$) are measured using a traveling microscope.

   • The squared diameter for a dark ring in air is given by $D_n^2=4nR\lambda$.
   • The difference between the squares of the diameters provides: $D_{(n+p),\text{air}}^2-D_{n,\text{air}}^2=4pR\lambda \dots (1)$

2. Measurement in Liquid: The liquid whose refractive index ($\mu$) is to be found is introduced between the lens and the plate, displacing the air film. The diameters of the same $n^{th}$ dark ring ($D_{n,\text{liquid}}$) and $(n+p{)}^{th}$ dark ring ($D_{(n+p),\text{liquid}}$) are measured again.

   • The diameter for a dark ring formed in a liquid of refractive index $\mu$ is given by $D_n^2=\frac{4nR\lambda }{\mu }$.
   • The difference between the squares of the diameters in the liquid is: $D_{(n+p),\text{liquid}}^2-D_{n,\text{liquid}}^2=\frac{4pR\lambda }{\mu }\dots (2)$

3. Calculation: By dividing Equation (1) by Equation (2), the constant parameters ($4pR\lambda$) are cancelled out, leaving the ratio dependent solely on the diameters measured and the refractive index: $\frac{D_{(n+p),\text{air}}^2-D_{n,\text{air}}^2}{D_{(n+p),\text{liquid}}^2-D_{n,\text{liquid}}^2}=\frac{4pR\lambda }{\frac{4pR\lambda }{\mu }}=\mu$ Thus, the refractive index of the liquid is calculated using the formula: $\mu =\frac{{\mathbf{D}}_{(\mathbf{n}+\mathbf{p}),\text{air}}^2-{\mathbf{D}}_{\mathbf{n},\text{air}}^2}{{\mathbf{D}}_{(\mathbf{n}+\mathbf{p}),\text{liquid}}^2-{\mathbf{D}}_{\mathbf{n},\text{liquid}}^2}$.

47. Explain why in the Newton’s rings, the rings get closer in radially outward directions.

The rings in the Newton’s ring pattern are unequally spaced and become closer together as the radius increases. This is explained by the relationship between the diameter ($D$) of a dark ring and its serial number ($n$).

1. Diameter Dependence: For the dark rings in reflected light, the squared diameter ($D_n^2$) is directly proportional to the serial number ($n$) and the product of the radius of curvature ($R$) and wavelength ($\lambda$): $D_n^2=4nR\lambda$ This means the diameter itself is proportional to the square root of the natural number $n$: ${\mathbf{D}}_{\mathbf{n}}\propto \sqrt{\mathbf{n}}$.

2. Spacing Calculation: The spacing between two consecutive dark rings is the difference between their diameters: $S_n=D_{n+1}-D_n$. $S_n=\sqrt{4(n+1)R\lambda }-\sqrt{4nR\lambda }=\sqrt{4R\lambda }(\sqrt{n+1}-\sqrt{n})$

3. Decreasing Separation: As $n$ increases (moving radially outward), the term $(\sqrt{\mathbf{n}+1}-\sqrt{\mathbf{n}})$ decreases. For example, the difference between $\sqrt{2}$ and $\sqrt{1}$ is larger than the difference between $\sqrt{10}$ and $\sqrt{9}$. Since the spacing between rings decreases continuously as $n$ increases, the rings appear to get closer and closer as they move outward from the center.

48. A parallel beam of light ($\lambda =5870\text{ A}$) is incident on a thin glass plate ($\mu =1.5$), such that the angle of refraction into the the plate is $60^{\circ }$. Calculate the smallest thickness of the glass plate which will appear dark by reflection.

To appear dark by reflection, the condition for minimum intensity must be satisfied.

Given:

• Wavelength, $\lambda =5870\text{ A}=5870\times 10^{-8}\text{ cm}$
• Refractive Index, $\mu =1.5$
• Angle of Refraction, $r=60^{\circ }$

Formula for Minima (Dark fringe) in Reflected Light: The condition for destructive interference (darkness) in a thin film is: $2\mu t\cos r=n\lambda$ We need to find the smallest thickness ($t$). This corresponds to setting the order $n=1$ (since $n=0$ implies $t=0$).

Calculation: Solving for $t$ with $n=1$: $t=\frac{1\times \lambda }{2\mu \cos r}$ $t=\frac{5870\times 10^{-8}\text{ cm}}{2\times 1.5\times \cos (60^{\circ })}$ Since $\cos (60^{\circ })=0.5$: $t=\frac{5870\times 10^{-8}\text{ cm}}{2\times 1.5\times 0.5}$ $t=\frac{5870\times 10^{-8}\text{ cm}}{1.5}$ $t=3913.33\times 10^{-8}\text{ cm}$

The smallest thickness of the glass plate which will appear dark by reflection is approximately $3913\text{ A}$ or $3913.33\times 10^{-8}\text{ cm}$.

48. In Newton’s ring experiment the diameter of 5th dark is $0.65\text{ cm}$ and that 15th dark ring is $0.95\text{ cm}$. The wavelength of source is used $6000\text{ A}$. Calculate the radius of curvature of a convex surface of a lens in contact with the glass plate.

We use the formula derived from the Newton’s dark ring condition for determining the radius of curvature ($R$).

Given:

• $n=5$, $D_5=0.65\text{ cm}$
• $n+p=15$, $D_{15}=0.95\text{ cm}$
• Wavelength, $\lambda =6000\text{ A}=6000\times 10^{-8}\text{ cm}$

Determine p: $p=(n+p)-n=15-5=10$

Formula for Radius of Curvature ($R$): $R=\frac{D_{(n+p)}^2-D_n^2}{4p\lambda }$

Calculation: $R=\frac{(0.95{)}^2-(0.65{)}^2}{4\times 10\times (6000\times 10^{-8}\text{ cm})}$

1. Calculate the difference in squared diameters: $D_{15}^2-D_5^2=0.9025-0.4225=0.48{\text{ cm}}^2$

2. Calculate the denominator term ($4p\lambda$): $4p\lambda =4\times 10\times 6000\times 10^{-8}\text{ cm}$ $4p\lambda =240,000\times 10^{-8}\text{ cm}=0.0024\text{ cm}$

3. Calculate $R$: $R=\frac{0.48{\text{ cm}}^2}{0.0024\text{ cm}}=200\text{ cm}$

The radius of curvature of the convex surface is $200\text{ cm}$.