Question

A convex lens of focal length \(5\,\text{cm}\) and a concave lens of focal length \(4\,\text{cm}\) are placed in contact and a point object is placed at \(10\,\text{cm}\) from the system. In this arrangement magnification is \(m_1\). Now keeping the system as it is, the concave lens is moved \(1\,\text{cm}\) away and the magnification becomes \(m_2\). Find \( \dfrac{m_1}{m_2} \).

• A. \( \dfrac{5}{6} \)
• B. \( \dfrac{4}{7} \)
• C. \( 6 \)
• D. \( 7 \)

Hint:

For multi-lens systems:
If lenses are separated, treat image of first lens as object for the next
Overall magnification is the product of individual magnifications

Answer 1

The Correct Option is A

Analysis: Stoichiometry: 6 moles HCl $\rightarrow$ 3 moles H$_2$. Ratio: 1 mole HCl produces 0.5 moles H$_2$. Step 1: Calculate Volume. At STP (1 atm, 273 K), 1 mole gas = 22.4 L. 0.5 moles H$_2$ = \(0.5 \times 22.4 = 11.2\,\text{L}\). Conclusion: Option (2) states exactly this.