Summary of HC Verma Solutions Part 1 Chapter 19: Optical Instruments
FAQs on HC Verma Solutions Class 11 Chapter 19 - Optical Instruments
1. What is the correct method for solving problems on compound microscopes from HC Verma Class 11 Chapter 19?
To solve problems on compound microscopes from HC Verma, follow this step-by-step method:
First, apply the lens formula (1/v - 1/u = 1/f) to the objective lens to find the position and nature of the intermediate image.
Treat this intermediate image as the object for the eyepiece lens.
Apply the lens formula again for the eyepiece to determine the final image position. This is often at infinity (for normal adjustment) or at the near point (D).
Calculate the total magnifying power by multiplying the magnification of the objective and the eyepiece (M = m_o × m_e).
Always use the Cartesian sign convention meticulously, as it is crucial for accurate results in HC Verma problems.
2. How do you calculate the magnifying power of an astronomical telescope when the final image is at the near point, as per HC Verma solutions?
In HC Verma problems, when the final image from an astronomical telescope is formed at the near point (D), it provides maximum magnification. The formula used for this specific case is:
M = -f₀/fₑ (1 + fₑ/D)
Here, f₀ is the focal length of the objective lens, fₑ is the focal length of the eyepiece, and D is the least distance of distinct vision (typically 25 cm). The negative sign indicates that the final image is inverted.
3. Why do HC Verma solutions for telescopes often solve for two cases: final image at infinity and at the near point?
HC Verma solutions explore both cases to teach the full operational range of a telescope. These two adjustments represent:
Final Image at Infinity (Normal Adjustment): This is the most comfortable viewing position for the human eye as the eye is completely relaxed. It provides the minimum magnification but allows for prolonged observation without strain.
Final Image at Near Point (D): This adjustment provides the maximum possible magnification the instrument can offer. However, it requires the eye to be fully strained, making it unsuitable for long viewing sessions. Solving for both highlights the trade-off between magnification and viewer comfort.
4. What is a common mistake when solving HC Verma problems on correcting vision defects like myopia and hypermetropia?
A common mistake is the incorrect application of sign conventions for lens power and distances. Remember:
For myopia (nearsightedness), the corrective lens is a concave lens, which has a negative focal length and negative power. The lens must form the image of an object at infinity (u = ∞) at the person's defective far point (v = -x).
For hypermetropia (farsightedness), the corrective lens is a convex lens, having a positive focal length and positive power. The lens must form the image of an object at the normal near point (u = -25 cm) at the person's defective near point (v = -y).
Confusing these signs is the most frequent source of error in solutions.
5. How are concepts from Geometrical Optics applied in the HC Verma solutions for Optical Instruments?
The chapter on Optical Instruments is a direct application of principles from Geometrical Optics. The solutions for microscopes and telescopes in HC Verma are built upon these fundamental concepts:
The Lens Formula (1/v - 1/u = 1/f) is used for each lens (objective and eyepiece) to locate images.
The formula for Linear Magnification (m = v/u) is essential to determine the size of the intermediate and final images.
A strict and consistent use of Cartesian sign convention is non-negotiable for solving the complex multi-lens systems found in this chapter.
Essentially, an optical instrument problem is broken down into a series of simpler single-lens problems.
6. How does the solution for a terrestrial telescope differ from an astronomical telescope in HC Verma's problems?
The key difference is the orientation of the final image. An astronomical telescope produces an inverted image, which is fine for viewing stars. A terrestrial telescope, used for viewing objects on Earth, requires an upright image. To achieve this, the solution methodology incorporates an extra component:
An erecting lens (or a prism system) is placed between the objective and the eyepiece.
This lens takes the inverted intermediate image from the objective and inverts it again, producing an upright image. This upright image then acts as the object for the eyepiece.
This adds an extra calculation step and increases the overall length of the telescope compared to an astronomical one of similar power.
7. Where can I find accurate, step-by-step HC Verma Solutions for Class 11 Physics Chapter 19 (Optical Instruments)?
You can find detailed and reliable solutions for HC Verma's 'Concepts of Physics' Class 11 Chapter 19 on Vedantu. The solutions are crafted by subject-matter experts, ensuring each problem is solved with a clear, step-by-step approach that is easy to understand. These solutions strictly follow the problem-solving logic and conceptual depth required for the 2025-26 academic session, helping you build a strong foundation in optical instruments.
