In a compound microscope, the objective lens forms a real, inverted, and diminished image at the focal plane of the eyepiece. The eyepiece acts as a magnifier to form a virtual, erect, and magnified image at the least distance of distinct vision. The magnifying power of the compound microscope is given by the product of the magnifying powers of the objective lens and the eyepiece lens : The magnifying power of the objective lens is given by: where is the image distance and is the object distance for the objective lens. Since the image is formed at the focal length of the objective lens , we have: For the eyepiece, the magnifying power is given by: where is the least distance of distinct vision and is the focal length of the eyepiece. Thus, the total magnifying power is:

Using Snell’s law for the critical angle : Hence, Thus, the critical angle is: Thus, the correct answer is:

For the given case, using Snell’s law: Here, since the refracted ray is parallel to the interface, , and hence: Thus, the refractive index of medium B is: Since the refractive index of the denser medium (A) is , we have: Thus, the correct answer is:

Question: The focal length of a concave mirror in air is . When the mirror is immersed in a liquid of refractive index , its focal length will become:

1. Key Concept:

The focal length of a concave mirror does not depend on the surrounding medium because the reflection from a mirror depends only on its radius of curvature and not on the refractive index of the medium. However, this is true only for reflection-based optics. In some competitive exams, the question may involve interpreting the apparent change in focal length due to refraction effects surrounding the mirror (such as using the mirror inside a liquid).

In such cases, the formula used is:

where is the refractive index of the surrounding medium with respect to air.

Here, , so:

2. Final Answer:

Option (A) is correct.

Optical fibers transmit light signals using the principle of total internal reflection. The light signals enter the fiber at an angle greater than the critical angle, which causes the light to be reflected entirely within the fiber. This continuous reflection ensures that the light travels through the fiber even if the fiber is bent. 

Step 1: Finding the image formed by the first lens . For lens A, the object distance is (since the object is real) and the focal length is . Using the lens formula: So, the image formed by lens A is at , which is real and inverted.

Step 2: Finding the image formed by the second lens . The image formed by lens A acts as the object for lens B. The object distance for lens B is the distance between the two lenses, i.e., . Using the lens formula for lens B: Thus, the final image is formed at a distance of 12.5 cm from lens B, which is real and inverted.

In this question, the arrangement described suggests that a real image is formed by a lens or optical system where the object consists of a vertical arrow and a horizontal line with a ball.
The characteristics of the image formed are:
- Since the image is real, it will be inverted.
- The image should retain the relative orientation of the object but will be mirrored vertically, as real images formed by a lens are typically inverted.
Therefore, the correct answer corresponds to the image formed where the image is inverted and real.

When a parallel beam of light enters a water surface obliquely, the beam undergoes refraction due to the change in the speed of light when passing from one medium (air) to another (water). Let's break down the effect of this refraction on the width of the beam:

1. Refraction of Light at the Water Surface:

When light passes from air (where the refractive index is approximately 1) into water (with a refractive index of about 1.33), the change in speed causes the light to bend according to Snell's law. The refractive index ( ) is related to the angle of incidence ( ) and the angle of refraction ( ) as:

• is the refractive index of air (1),
• is the refractive index of water (1.33),
• is the angle of incidence in air, and
• is the angle of refraction in water.

2. Effect on the Width of the Beam:

When the light enters the water obliquely, the beam bends, causing the light rays to spread out more (or less) depending on the direction of the incident light. Since the speed of light in water is slower, the light rays are refracted towards the normal, causing the beam to become narrower in the direction of propagation.

• As the beam enters the water, the width of the beam decreases due to the bending of the rays towards the normal. This narrowing of the beam occurs because the light slows down in the water, and the rays are closer together after refraction.
• The effect is more pronounced when the angle of incidence is large, as the light bends more sharply towards the normal. For a smaller angle of incidence, the change in the width is less noticeable.

3. Conclusion:

When a parallel beam of light enters the water surface obliquely, the width of the beam decreases due to the refraction of light as it slows down and bends towards the normal. The greater the angle of incidence, the more pronounced the narrowing of the beam.

Calculation of Focal Length of Concave Lens

Given:

• Object distance from convex lens,
• Focal length of convex lens,
• Shift in image position with concave lens added,

Step 1: Use Lens Formula for Convex Lens

The lens formula is:

Substituting the given values for and :

So, the image distance is:

The initial image distance is (since the image formed is real, we use object-image symmetry).

Step 2: Adding the Concave Lens

After adding the concave lens, the image shifts by behind the original screen. So, the new image distance becomes:

Step 3: Effective Focal Length of the System

Since the lenses are in contact, the effective focal length of the system can be calculated using the lens formula:

Now, use the lens formula again for the system:

Substituting the values and :

Hence, the effective focal length is .

Step 4: Focal Length of the Concave Lens

Now, we can solve for the focal length of the concave lens: Substituting the known values: Solving for : Therefore, the focal length of the concave lens is:

Final Answer:

The focal length of the concave lens is .

Phasor Diagrams, Resonance Condition, and Resonant Frequency

(I) Phasor Diagrams

In AC circuits, the relationship between voltage and current can be represented using phasor diagrams:

For Capacitor (X):

• In a capacitor, the voltage leads the current by . This is because the current in a capacitor leads the voltage due to the nature of capacitive reactance.
• The phasor diagram shows that the phasor for voltage is ahead of the phasor for current.

For Inductor (Y):

• In an inductor, the voltage lags behind the current by . This happens because the voltage across an inductor opposes the change in current (inductive reactance).
• The phasor diagram shows that the phasor for voltage is behind the phasor for current.
  
  

(II) Condition of Resonance and Resonant Frequency

When a capacitor and an inductor are connected in series with a resistor (forming an RLC circuit), resonance occurs under specific conditions. The condition of resonance is when the inductive reactance ( ) equals the capacitive reactance ( ).

Condition for Resonance:

The resonance condition is given by:

Solving for , we get:

Resonant Frequency:

The resonant frequency is given by:

Impedance at Resonance:

At resonance, the net reactance becomes zero because . Therefore, the total impedance in the circuit is purely resistive:

At Resonance:

• The impedance is equal to the resistance of the circuit.
• The current is maximum and is in phase with the voltage.

Summary:

Resonance in an RLC circuit occurs when the inductive reactance equals the capacitive reactance, leading to maximum current and a purely resistive impedance. The resonant frequency is given by:

We need to evaluate the given statements about magnification for different mirrors and lenses.

1. Understanding the Concepts:

Magnification: The magnification of an image is the ratio of the image's height to the object's height. It is given by the formula , where is the image height and is the object height. A positive magnification means the image is upright, and a negative magnification means the image is inverted.

2. Analyzing the Statements:

Statement (A): For a convex mirror, magnification is always negative.
This statement is incorrect. For a convex mirror, the image formed is always virtual, upright, and smaller than the object. The magnification for a convex mirror is always positive, not negative.

Statement (B): For all virtual images formed by a mirror, magnification is positive.
This statement is correct. Virtual images formed by mirrors, such as in concave mirrors or convex mirrors, are always upright, resulting in a positive magnification.

Statement (C): For a concave lens, magnification is always positive.
This statement is correct. A concave lens always forms a virtual, upright image, which results in a positive magnification.

Statement (D): For real and inverted images, magnification is always negative.
This statement is correct. Real and inverted images, which are formed by concave mirrors or convex lenses, always have a negative magnification.

3. Conclusion:

The incorrect statement is Statement (A): "For a convex mirror, magnification is always negative." This is false because the magnification for convex mirrors is always positive.

Final Answer:

The incorrect statement is (A) For a convex mirror, magnification is always negative.

When light travels from one medium to another (such as from water to air), the change in speed causes the light rays to bend at the interface. This bending of light makes the straw appear broken or bent at the surface of water. The light rays from the part of the straw submerged in water are refracted at the water-air interface, making the submerged part appear displaced from the rest of the straw. 

The light travels along the surface of the cylindrical rod. For this to happen, the angle of incidence must be such that the light is refracted along the surface. Using Snell's law at the interface between the rod and air: Given and , we have: Thus: Therefore, the angle is .

To determine the apparent depth of a coin placed at the bottom of a beaker filled with water, we use the concept of the refractive index and Snell's law. The apparent depth ( ) is given by the formula:

where is the actual depth of the coin and is the refractive index of the medium (water in this case).

Given:

• Actual depth,
• Refractive index of water,

Substituting the values, we find:

To solve this, multiply by the reciprocal of :

Therefore, the depth of the coin, when viewed along the near normal direction, is .