In a convex mirror having radius of curvature 30 cm the height of image is half the object height. What will be the object (in cm) distance?

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alok kumar singh | Contributor-Level 10

4 weeks ago

f = 15

$m = - \frac{v}{u} = + \frac{1}{2}$

$v = - \frac{u}{2}$

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{2}{- u} + \frac{1}{u} = \frac{1}{f}$

u = – f = – 15cm

f = 15 <math> <mrow> <mi>m</mi> <mo>=</mo> <mo>−</mo> <mfrac> <mrow> <mi>v</mi> </mrow> <mrow> <mi>u</mi> </mrow> </mfrac> <mo>=</mo> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>            <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1820663765"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>v</mi> <mo>=</mo> <mo>−</mo> <mfrac> <mrow> <mi>u</mi> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math>           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1820663766"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>v</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>u</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>f</mi> </mrow> </mfrac> </mrow> </math>            <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1820663767"> </o:OLEObject></xml><! [endif]-> <math> <mrow> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mo>−</mo> <mi>u</mi> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>u</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mi>f</mi> </mrow> </mfrac> </mrow> </math>            <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1820663768"> </o:OLEObject></xml><! [endif]->u = – f = – 15cm

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In a convex mirror having radius of curvature 30 cm the height of image is half the object height. What will be the object (in cm) distance?

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