An object placed in front of a concave mirror of focal length 0.15 m produces a virtual image which is twice the size of the object. The position of the object from the mirror as per sign convention is
A
−6.5 cm
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B
+6.5 cm
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C
−7.5 cm
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D
+7.5 cm
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Solution
The correct option is C−7.5 cm
Here, f=−0.15 m and m=+2 as mirror is concave and image is virtual. We know, m=ff−u ⇒+2=−0.15−0.15−u ⇒−0.3−2u=−0.15 ⇒2u=−0.15 ⇒u=−0.152=−0.075 m=−7.5 cm