SATHEE: Optics 3 Question 15

Optics 3 Question 15

15. The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is

(2006, 3M)

(a) $5 \pm 0.1$

(b) $5 \pm 0.05$

(d) $0.5 \pm 0.05$

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Answer:

Correct Answer: 15. (b)

Solution:

From the lens formula,

$\begin{aligned} \frac{1}{f} & = \frac{1}{v} - \frac{1}{u} we have, \\ \frac{1}{f} & = \frac{1}{10} - \frac{1}{- 10} \\ or & f u r t h e r, & = + 5 \\ and Δ u & = 0.1 \\ Δ v & = 0.1 (from the graph) \end{aligned}$

Now, differentiating the lens formula, we have

$or \begin{aligned} \frac{Δ f}{f^{2}} & = \frac{Δ v}{v^{2}} + \frac{Δ u}{u^{2}} \\ Δ f & = \frac{Δ v}{v^{2}} + \frac{Δ u}{u^{2}} f^{2} \end{aligned}$

Substituting the values, we have

$\begin{aligned} Δ f & = \frac{0.1}{10^{2}} + \frac{0.1}{10^{2}} (5)^{2} = 0.05 \\ ∴ f \pm Δ f & = 5 \pm 0.05 \end{aligned}$

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Optics 3 Question 15

15. The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics 3 Question 15

15. The graph between object distance u and image distance v for a lens is given below. The focal length of the lens is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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15. The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is