Match List - I with List - II.
| List – I | List – II | ||
|---|---|---|---|
| (A) | Lens formula | (I) | \( \displaystyle\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \) |
| (B) | Magnifying power of Microscope | (II) | \( f_0/f_e \) |
| (C) | Magnifying power of telescope | (III) | \( \displaystyle\frac{L}{f_0}\times\frac{D}{f_e} \) |
| (D) | Mirror Equation | (IV) | \( \displaystyle\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(A) Lens formula — this matches (IV) \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \). This is the thin lens formula in the Cartesian sign convention.
(B) Magnifying power of Microscope — this matches (III) \( \frac{L}{f_0}\times\frac{D}{f_e} \). A compound microscope has \( m = m_om_e = \frac{L}{f_0}\times\frac{D}{f_e} \), with L the tube length and D = 25 cm.
(C) Magnifying power of telescope — this matches (II) \( f_0/f_e \). For a telescope in normal adjustment \( m = f_0/f_e \), the ratio of the focal lengths of the objective and the eyepiece.
(D) Mirror Equation — this matches (I) \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \). This is the mirror formula in the Cartesian sign convention.
NCERT (2026-27):
Refraction by a Lens – “… Applying the sign convention, BO = –u, DI = +v, we get
\( \displaystyle \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \qquad (9.23) \)
Equation (9.23) is the familiar thin lens formula.”
The Mirror Equation – “… Dividing it by v, we get
\( \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \qquad (9.7) \)
This relation is known as the mirror equation.”
The Microscope – “… Thus, the total magnification [(according to Eq. (9.33)], when the image is formed at infinity, is
\( \displaystyle m = m_o m_e = \left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right) \qquad (9.45) \)”
Telescope – “… The magnifying power m is the ratio of the angle \( \beta \) subtended at the eye by the final image to the angle \( \alpha \) which the object subtends at the lens or the eye. Hence
\( \displaystyle m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e}\cdot\frac{f_o}{h} = \frac{f_o}{f_e} \qquad (9.46) \)”