एक वस्तु f = 10 text{ cm} फोकस दूरी वाले उत्त | vedclass

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(B) दिया गया है: $f = +10 	ext{ cm}$,$u = -30 	ext{ cm}$,$V_O = +4 	ext{ cm/sec}$,$V_L = +2 	ext{ cm/sec}$.लेंस सूत्र का उपयोग करते हुए: $\frac{1}

Vedclass · Accepted Answer

$2.5 	ext{ cm/sec}$

Vedclass · Answer

(B) दिया गया है: $f = +10 	ext{ cm}$,$u = -30 	ext{ cm}$,$V_O = +4 	ext{ cm/sec}$,$V_L = +2 	ext{ cm/sec}$.लेंस सूत्र का उपयोग करते हुए: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.समय के सापेक्ष अवकलन करने पर: $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$,जहाँ $\frac{dv}{dt} = V_{I/L}$ और $\frac{du}{dt} = V_{O/L}$.$V_{I/L} = \left(\frac{v}{u}ight)^2 V_{O/L} = m^2 V_{O/L}$.सबसे पहले,$v$ ज्ञात करें: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15} \implies v = +15 	ext{ cm}$.आवर्धन $m = \frac{v}{u} = \frac{15}{-30} = -0.5$.सापेक्ष वेग: $V_{O/L} = V_O - V_L = 4 - 2 = +2 	ext{ cm/sec}$.$V_{I/L} = m^2 V_{O/L} = (-0.5)^2 	imes 2 = 0.25 	imes 2 = 0.5 	ext{ cm/sec}$.चूंकि $V_{I/L} = V_I - V_L$,इसलिए $V_I = V_{I/L} + V_L = 0.5 + 2 = 2.5 	ext{ cm/sec}$.

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