Optics

• Refractive Index: The refractive index of a material determines how much light bends when it enters that material from another medium. The lens is visible in the air because its refractive index is different from that of the air, creating refraction and thus, visibility.
• When Immersed in Liquid: If the lens and the liquid have the same refractive index, light doesn't bend when passing through the lens. This lack of bending or refraction means there's no change in the path of light, so the lens becomes invisible.

Step 1: Trace the path of the ray of light through the prism. When a ray of light enters a prism with refractive index greater than that of the surrounding medium (typically air), it bends towards the normal at the first interface (at angle of incidence ), passes through the prism, and bends away from the normal at the second interface (at angle of exit ). The prism has an apex angle .

Step 2: Derive the formula for the angle of deviation ( ). The angle of deviation ( ) is the angle by which the light ray deviates from its original direction after passing through the prism. It can be expressed in terms of the angle of incidence ( ), the angle of exit ( ), and the prism's apex angle ( ) as follows: This relationship arises because the external deviation is equal to the sum of the angles of incidence and emergence minus the angle of the prism.

Step 3: Graph the variation of with . The relationship between and is typically non-linear, showing that decreases with an increase in up to a minimum value (at the minimum deviation condition) and then increases. The graph of versus will have a "U" shape, indicating the minimum deviation occurs when the light ray passes symmetrically through the prism.

Given: Radius of , Radius of , Permeability of free space, The mutual inductance between two coaxial circular loops is given by: Substitute the given values: Simplify the expression: Calculate the denominator: Substitute : Therefore, the mutual inductance of the loops is:

Image Formation by a Convex Lens and Concave Mirror

Step 1: Image Formation by the Convex Lens
- A parallel beam of light incident on a convex lens converges at the focus of the lens.
- Given focal length of the convex lens: - Since parallel rays converge at the focal point, the image formed by the convex lens is: - This image acts as the object for the concave mirror.

Step 2: Object Distance for the Concave Mirror
- Distance between the lens and the mirror: - Distance of the image formed by the lens from the mirror: - Focal length of the concave mirror: Using the mirror formula:

Step 3: Final Image Position
- The negative sign indicates that the final image is on the same side as the mirror.
- Distance from the mirror: 10 cm.
- Since the mirror is 40 cm from the lens, the final image is: - Since the image distance from the lens is greater than the focal length, the final image is real and inverted.

Step 4: Conclusion
- Final image is formed at 10 cm to the right of the lens.
- Thus, the correct answer is:

Power of a Combination of Lenses

Step 1: Understanding Power of Lenses
- The power of a lens is given by: where: - = Focal length in cm.

Step 2: Calculating the Power of Each Lens
- Converging lens (convex): - Diverging lens (concave):

Step 3: Total Power of the Combination
Since the lenses are in contact, the net power is:

Step 4: Conclusion
Thus, the power of the combination is: which matches option (B).

The focal length of a lens in a medium is given by the lens maker's formula: For a double convex lens, and . The refractive index of glass and water . Substituting these values: Thus, the focal length of the lens in water is .

Given Data (for quick reference)

• Mediums: Air (n1=1) → Glass (n2=1.5) → Air.
• Sphere (ball lens) radius: R = 15 cm; diameter: D = 30 cm.
• Incoming beam: parallel (object at infinity).
• Sign convention (Cartesian): distances measured from each surface; positive to the right (direction of light).
• Refracting-surface formula:

n₂/v − n₁/u = (n₂ − n₁)/R

Step-by-Step Derivation

Step 1 — First refraction (air → glass) at the left surface

Object at infinity ⇒ 1/u₁ ≈ 0. For the left surface, the center of curvature is to the right, so R₁ = +15 cm.

n₂/v₁ − n₁/u₁ = (n₂ − n₁)/R₁ ⇒ (1.5)/v₁ − 0 = (1.5 − 1)/15 ⇒ v₁ = 45 cm.

This means the rays would meet 45 cm to the right of the left surface (inside the glass) if there were no second surface.

Step 2 — Geometry for the second surface

The sphere’s thickness along the axis is its diameter: 30 cm. The “would-be” focus of Step 1 is therefore 45 − 30 = 15 cm to the right of the right surface. For the right surface, this is a virtual object located u₂ = +15 cm to its right (our positive direction remains to the right).

Step 3 — Second refraction (glass → air) at the right surface

Here, the center of curvature is to the left, so R₂ = −15 cm, with n₁ = 1.5 (inside) and n₂ = 1 (outside).

n₂/v₂ − n₁/u₂ = (n₂ − n₁)/R₂
1/v₂ − (1.5/15) = (1 − 1.5)/(−15) ⇒ 1/v₂ − 0.1 = 1/30 ⇒ 1/v₂ = 0.1333… ⇒ v₂ = 7.5 cm.

Final answer: The emergent rays form a real image on the axis at a distance 7.5 cm to the right of the sphere’s right surface (i.e., outside the ball).
Equivalently, the ball lens has:

• Effective focal length (EFL) from the center: f = nD / (4(n−1)) = (1.5×30)/(4×0.5) = 22.5 cm.
• Back focal length (BFL) from the right surface: BFL = f − R = 22.5 − 15 = 7.5 cm.

Key Takeaways (Exam-ready)

• For parallel light, treat each surface with n₂/v − n₁/u = (n₂ − n₁)/R and carry the intermediate image to the next surface.
• Ball-lens quick check: EFL = nD / (4(n−1)), BFL = EFL − R. For n=1.5, R=15 cm, BFL=7.5 cm.
• The final image here is real, highly reduced (a point focus), located outside the sphere.

For a spherical surface, the lens-maker's equation is given by: where and are the refractive indices of the medium on either side of the surface, is the radius of curvature of the surface, and is the radius of curvature of the second surface. In this case, we are dealing with a spherical convex surface, so (since it is an open surface), and the equation simplifies to: Substitute and : Thus, the focal length is: Since the object is placed at a distance from the surface, we can use the lens formula: where (object distance is negative), and . Substituting the values: Simplifying: Thus, the image distance is: The negative sign indicates that the image is virtual, formed on the same side as the object. Therefore, the image is virtual, formed at a distance behind the surface.

The light from the point source will emerge out of the liquid surface within a circle of radius due to total internal reflection. The critical angle for total internal reflection is given by: The radius of the circle on the surface is: Using , we get: The area of the circle is: Thus, the correct answer is (C).

When a plane wave travels from one medium to another, its direction of propagation changes due to the difference in the speed of light in the two media. This phenomenon is called refraction. The angle of incidence in the denser medium is related to the angle of refraction in the rarer medium by Snell’s law.

Diagram for Refraction:

Below is the diagram showing the refraction of a plane wave at the boundary between two media:

Incident plane wave in denser medium

↓

Angle of incidence (i)

Refraction occurs at the boundary

↓

Angle of refraction (r) in the rarer medium

Snell’s Law:

Snell’s law relates the angle of incidence i and the angle of refraction r to the refractive indices of the two media. It is given by:

Snell's Law Formula:

where:

• n1 is the refractive index of the denser medium,
• n2 is the refractive index of the rarer medium,
• i is the angle of incidence,
• r is the angle of refraction.

This equation can be derived using the principles of wave theory and the relationship between the speed of light in the two media. The refractive index n is defined as:

Refractive Index Formula:

where c is the speed of light in vacuum and v is the speed of light in the medium.

Thus, Snell’s law ensures that the ratio of the sine of the angles is equal to the ratio of the refractive indices.

Final Answer:

The refraction of the plane wave can be described by Snell’s law, and the relationship between the angles of incidence and refraction in two media is given by:

Snell's Law:

A wavefront of a light wave is a surface on which the phase of the wave is constant. It is a surface where every point on it has the same phase of oscillation.

- In the case of plane waves, the wavefronts are flat and the light rays are parallel to each other.

- For spherical waves, the wavefronts are spherical and radiate outward from a point source.

Refraction of a Plane Wavefront through a Convex Lens:

When a plane wavefront strikes a convex lens, the light rays are refracted (bent) due to the change in speed as they move from one medium to another. The convex lens converges the parallel light rays to a point called the focal point.

Shape of the Refracted Wavefront:

• Before refraction, the wavefront is a straight line (plane wave).
• After passing through the convex lens, the wavefronts bend and become curved.
• The refracted wavefronts now resemble part of a spherical wavefront. The light rays converge towards the focal point of the lens.

The refracted wavefronts are no longer parallel but are bent inward, becoming part of a spherical surface centered at the focal point of the lens.

Final Answer:

The plane wavefront, after passing through the convex lens, becomes curved and converges toward the focal point. The shape of the refracted wavefront is part of a spherical wavefront.

To find the focal length of the lens, we use the lens formula:

where is the focal length, is the image distance, and is the object distance. Given that the object distance cm (convention: object distance is negative) and the screen is 50 cm away from the object, the image distance cm.

Substitute these values into the lens formula:

Convert to a common denominator:

Therefore,

So, the focal length of the lens is 16 cm.

The intensity in Young’s double slit experiment is given by:

Where:

- is the intensity of the central maximum,

- is the path difference between the two waves,

- is the wavelength of the light.

Given: , we substitute into the equation:

Since , we get:

Final Answer:
The intensity at the point where the path difference is is:

A thin lens is a transparent optical medium with two spherical surfaces, and it can be treated as the combination of two spherical surfaces. The lens maker formula gives the focal length of the lens in terms of the refractive index , the radii of curvature and of the two surfaces, and the thickness of the lens (if required). The formula is: where:
is the focal length of the lens,
and are the radii of curvature of the first and second spherical surfaces of the lens, and
is the refractive index of the material of the lens.

For a thin lens, the lens formula is given by: where:
is the focal length,
is the image distance (distance from the lens to the image),
is the object distance (distance from the lens to the object).
The lens has two foci, one on each side, known as the first focal point and the second focal point.

To determine the new focal length when a convex lens is cut into two equal parts, we need to consider the effect of dividing the lens on its optical properties. The focal length of a lens is given by the lens maker's formula:

where is the refractive index and are the radii of curvature of the lens surfaces. However, when we physically cut a convex lens (which is initially symmetrical and thin) into two equal halves, the curvature and refractive index remain unchanged, but the aperture area is reduced.

For the new lens piece, the lens maintains its curvature properties but the diameter (or the aperture of the lens) is halved. This effectively changes the lens's ability to converge rays. A larger aperture allows better convergence due to less diffraction at the edges. Dividing the lens reduces the aperture, influencing the effective focal length of each piece.

For a lens cut along the principal axis, you essentially have a lens with half the aperture area, which increases the converging power given by:

This is because the portioned lens should act equivalently to the original smaller, complete lens with half the aperture area: focusing light to the same point but needing adjustment numerically due to geometry and optics context.

Therefore, the focal length of each half of the original lens is .

1. Intensity and Interference:

The intensity of the light at any point on the screen in a Young’s double-slit experiment depends on the interference of the two light waves reaching that point. The resultant intensity is given by the formula:

Where:

• and are the intensities of the individual waves.
• is the phase difference between the two waves.

Since both waves have the same intensity , we have . So, the expression for the resultant intensity becomes:

2. Phase Difference at Path Difference :

The phase difference is related to the path difference by the equation:

Given that the path difference , we substitute this into the equation for the phase difference:

3. Calculating the Intensity at the Point:

Now, substitute into the expression for the resultant intensity:

Since , we get:

4. Conclusion:

• The intensity at the point where the path difference is is .