Concave mirror numerical: u = -20 cm, f = -15 cm but I get v = -8.57 cm. Where is my sign going wrong?

7 answers & replies

• [+][–]Siddharth RaoDropperSample· 22d ago

  The mirror formula is

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

  so $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$. You added instead of subtracting. Correct:

  $\frac{1}{v} = -\frac{1}{15} + \frac{1}{20} = \frac{-4+3}{60} = -\frac{1}{60}$

  so $v = -60$ cm, i.e. the image is 60 cm in front of the mirror (real). Magnification is $m = -\dfrac{v}{u} = -\dfrac{-60}{-20} = -3$. Image is real, inverted and three times as large.

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• [+][–]Neel Chatterjee12thSample· 22d ago

  Sanity check: for a concave mirror, $f = 15$ cm and so $C$ is at $R = 30$ cm. Your object is at 20 cm, which lies between F (15) and C (30). Ray diagram rule: object between F and C gives a real, inverted, magnified image beyond C. So the image distance must be more than 30 cm. Your $-8.57$ cm is inside F, which is impossible for that case. Takes 10 seconds and saves marks.

  Quick table to memorise: object beyond C → image between F and C, smaller; at C → image at C, same size; between F and C → beyond C, bigger; at F → infinity; between P and F → virtual, erect, behind the mirror.

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• [+][–]Harsh Gupta12thSample· 22d ago

  Everyday mirrors help me remember. Dentist's and shaving mirrors are concave: when your face/tooth is between the pole and the focus you get a magnified, erect, virtual image. Torch and headlight reflectors are concave with the bulb at the focus, which sends out a parallel beam. Rear-view mirrors on bikes and cars are convex because they give a wider field of view, with smaller but always erect images. mirror ne sab theek kar diya 😎

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• [+][–]Nisha Verma12thSample· 22d ago

  On the magnification: it's $m = -\dfrac{v}{u}$ (equal to image height over object height), and the minus is there on purpose. With a real image, $u$ and $v$ are both negative so $v/u$ is positive, and the extra minus makes $m$ negative, which means inverted. In your corrected answer $m = -3$: real, inverted, three times as large. If $m$ came out +3, the image would be virtual and erect, which a concave mirror never gives for an object beyond the focus.

  Rule I use: negative $m$ → inverted → real. Positive $m$ → erect → virtual.

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  • [+][–]Harsh Gupta12thSample· 22d ago

    noted, the minus in m = -v/u is there on purpose. rear-view mirror one is my favourite example 😂

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• [+][–]Vivek Mishra12thSample· 22d ago

  For convex mirror: the focal length and radius of curvature are positive, because the focus is behind the mirror (on the side opposite to the incident light), and by the sign convention the distances on the side of incidence are negative while the opposite side is positive. Same formula $1/v + 1/u = 1/f$ works. The image is always virtual, erect and diminished, for any object position. For example $u = -20$, $f = +15$ gives $1/v = 1/15 + 1/20 = 7/60$, so $v = +8.57$ cm. Funny, your wrong number is the right answer for a convex mirror of the same focal length!

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  • [+][–]Riya Singh12thSample· 22d agoAuthor

    HA. my mistake wrote a poem about a different mirror. thank you all, the "object between F and C" check is going on a sticky note

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