A concave mirror has a focal length of 30.8 cm. The distance between an object and its image is 56.5 cm. Find?

A concave mirror has a focal length of 30.8 cm. The distance between an object and its image is 56.5 cm. Find (a) the object and (b) the image distance assuming that the object lies beyond the center of curvature.

(a): __________ cm

(b): __________ cm

2 Answers

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  • 9 years ago
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    Answers … (a) 100.8 cm … (b) 44.3 cm …

    … ( 1 / o ) + ( 1 / i ) = ( 1 / f ) … where … f = 30.8 cm …

    Since … f = R /2 … then … R = 2 f = 2 ( 30.8 cm ) = 61. 6 cm …

    so that … o > 61.6 cm … then … either ( o – i ) = 56.5 (case 1) …

    … or … ( i - o ) = 56.5 (case 2) … consider case 1 …

    … ( o – i ) = 56.5 … --> … i = o - 56.5 …

    … ( 1 / o ) + [ 1 / ( o – 56.5 ) ] = ( 1 / 30.8 ) …

    … { ( o – 56.5 ) + o } / [ o ( o – 56.5 ) ] = ( 1 / 30.8 ) …

    … ( 2 o – 56.5 ) ( 30.8 ) = o ( o – 56.5 ) …

    … ( 61.6 ) o – 1740.2 = o ² - (56.5 ) o …

    … o ² - ( 118.1) o + 1740.2 = 0… solve for o using quadratic

    formula … o = ½ { 118.1 ± √ [ ( - 118.1) ² - 4 (1) (1740.2 ) ] …

    … o₁ = 17.3 cm < 61.6 cm … o₂ = 100.8 cm > 61.6 cm is OK …

    … i = o₂ - 56.5 = 100.8 - 56.5 = 44.3 …

    For case 2 … ( i - o ) = 56.5 … --> … i = o + 56.5 …

    … ( 1 / o ) + [ 1 / ( o + 56.5 ) ] = ( 1 / 30.8 ) …

    … { ( o + 56.5 ) + o } / [ o ( o + 56.5 ) ] = ( 1 / 30.8 ) …

    … ( 2 o + 56.5 ) ( 30.8 ) = o ( o + 56.5 ) …

    … ( 61.6 ) o + 1740.2 = o ² + (56.5 ) o …

    … o ² - ( 5.1) o - 1740.2 = 0… solve for o using quadratic

    formula … o = ½ { 5.1 ± √ [ ( - 5.1) ² - 4 (1) ( -1740.2 ) ] …

    … o₁ = -39.24 cm … o₂ = 44.34 cm … both inadmissible solutions …

  • Anonymous
    4 years ago

    The radius of curvature is two times the focal length. So: f=R/2 f=120 cm/2 f=60 cm clean up for the gap of the image(di): a million/f = a million/do + a million/di di = fdo / (do-f) di = 60(20) / (20 - 60) di = 1200 / -40 di = - 30 cm clean up for the magnification(m): m = -di/do m = -(-30)/20 m = a million.5 ***the attitude of occurrence is the same because the attitude of mirrored image. So the attitude of mirrored image is 40 six ranges too.

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