Q4. Assertion: the objective of an astronomical telescope is made as large as possible. Reason: both the light-gathering power and the resolving power increase with the objective’s diameter. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion (NCERT page 245).
Q5. A person with a near point of 25 cm uses a compound microscope with an objective of focal length 8.0 mm and an eyepiece of 2.5 cm, and sees an object 9.0 mm from the objective with the final image at that 25 cm near point. Find the separation of the lenses and the magnifying power.
The objective gives vo = 1/(1/0.8 − 1/0.9) = 7.2 cm, a magnification of 8. For the final image at 25 cm, ue = −1/(1/25 + 1/2.5) ≈ −2.27 cm. The separation is 7.2 + 2.27 ≈ 9.47 cm, and m = 8 × (1 + 25/2.5) = 88 (NCERT Exercise 9.12, page 250; answer on page 346).
Q6. For the magnifier of Example 4 (f = 10 cm, as in NCERT’s answer), where should the card be held for the largest magnifying power, and what is it?
With the image at the near point, 1/u = 1/v − 1/f = −1/25 − 1/10, so u ≈ −7.14 cm; the magnification and the magnifying power are then both 25/7.14 = 3.5 (NCERT Exercise 9.23, page 251; answer on page 348).
Q7. With the same magnifier, where must the card be for each 1 mm2 square to appear 6.25 mm2 in area? Could you see it clearly with your eye close to the lens?
The linear magnification must be √6.25 = 2.5, so v = 2.5u, and 1/(2.5u) − 1/u = 1/10 gives u = −6 cm and |v| = 15 cm. The image is nearer than 25 cm, so the eye cannot focus on it clearly (NCERT Exercise 9.24, page 251; answer on page 348).