Question

12) the radius of a curvature of a concave mirror
15 cm and an object is placed at 7cm in front
of the mirror on principal axis the 'Where could
be Image formed.​

Answer 1

Answer:

Correct answer is: v = 14.12

Explanation:

R = 15 cm

U = -7cm

V = ?

F =R/2

F =15/2

mirror formula :- 1/V + 1/ U = 1/ F

= 1/V =1/F - 1/U

= 1/V =2/15 - 1/-7

take a LCM

= -7 + 15

105

= v= 14.12

I hope this answer will help you...

Answer 2

GiveN :

• Radius of curvature, r = 15 cm
• Object distance, u = 7 cm

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Note :

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In the case of Concave mirror, image will be formed on the same side of object . Hence, the object distance becomes positive.

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As we're given with the, radius of curvature & object distance we've to find the focal length of the mirror which is unknown to us.

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Also, if the radius of curvature is given, we can easily find out the focal length as focal length of a mirror is approximately half of its radius of curvature.

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$\because \:{ \boxed{\bf{r = 2f}}}$

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Substituting the values :

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$\longrightarrow\bf \: 15 = 2f$

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$\longrightarrow\bf \: f = \dfrac{15}{2} \: cm$

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Now, we've to calculate the image distance & we can find it by using the mirror formula. Mirror formula is given by :

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$\longrightarrow \bf \: \: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

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$\longrightarrow \bf \: \: \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$

where,

• v = image distance
• u = object distance
• f = focal length

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And we've the values,

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• v = unknown
• u = 7 cm
• f = $\sf \dfrac{15}{2}$

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Substituting the values :

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$\longrightarrow \bf \: \: \dfrac{1}{v} = \dfrac{1}{ \frac{15}{2} } - \dfrac{1}{7}$

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$\longrightarrow \bf \: \: \dfrac{1}{v} = \dfrac{2}{15 } - \dfrac{1}{7}$

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$\longrightarrow \bf \: \: \dfrac{1}{v} = \dfrac{14 - 15}{105}$

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$\longrightarrow \bf \: \: \dfrac{1}{v} = \dfrac{ -1}{105}$

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$\longrightarrow \large{ \boxed{\frak{\: v = - 105 \: cm}}}$

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$\therefore$ Nature of image is real and inverted with the image distance as -105 cm.