Distance between object and its image (magnified by ) is 30 cm. The focal length of the mirror used is , where magnitude of value of x is __________. [2025]

Question

Distance between object and its image (magnified by $- \frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $(\frac{x}{4}) cm$ , where magnitude of value of x is __________. [2025]

Smart Achievers · Accepted Answer

(45)

$m = \frac{f}{f - u} = - \frac{1}{3} \Rightarrow 4 f = u$

$\frac{1}{v} + \frac{1}{4 f} = \frac{1}{f} \Rightarrow v = \frac{4 f}{3}$

$(u - v) = \frac{8 f}{3} = 30$

$f = \frac{90}{8} = \frac{45}{4} \Rightarrow x = 45$

Solution

Q.
Distance between object and its image (magnified by $- \frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $(\frac{x}{4}) cm$ , where magnitude of value of x is __________. [2025]

Ans.
(45)

$m = \frac{f}{f - u} = - \frac{1}{3} \Rightarrow 4 f = u$

$\frac{1}{v} + \frac{1}{4 f} = \frac{1}{f} \Rightarrow v = \frac{4 f}{3}$

$(u - v) = \frac{8 f}{3} = 30$

$f = \frac{90}{8} = \frac{45}{4} \Rightarrow x = 45$

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Solution

Q. Distance between object and its image (magnified by –13) is 30 cm. The focal length of the mirror used is (x4) cm, where magnitude of value of x is __________. [2025]

Ans. (45) m=ff–u=–13 ⇒ 4f=u 1v+14f=1f ⇒ v=4f3 (u–v)=8f3=30 f=908=454 ⇒x=45

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Q.
Distance between object and its image (magnified by $- \frac{1}{3}$ ) is 30 cm. The focal length of the mirror used is $(\frac{x}{4}) cm$ , where magnitude of value of x is __________. [2025]

Ans.
(45)

$m = \frac{f}{f - u} = - \frac{1}{3} \Rightarrow 4 f = u$

$\frac{1}{v} + \frac{1}{4 f} = \frac{1}{f} \Rightarrow v = \frac{4 f}{3}$

$(u - v) = \frac{8 f}{3} = 30$

$f = \frac{90}{8} = \frac{45}{4} \Rightarrow x = 45$