I need a justified solution to the question above. Thanks. A converging lens has

Question

I need a justified solution to the question above. Thanks.

A converging lens has a focal length of 10.0 cm. Locale the object if a real image is located at a distance from the lens of (a) 20.0 cm and (b) 50.0 cm. What If? Redo the calculations if the images are virtual and located at a distance from the lens of (c) 20.0 cm and (d) 50.0 cm. How to justify the following answers to the question? (a) 20.0 cm from the lens on the front side (b) 12.5 cm from the lens on the front side (c) 6.67 cm from the lens on the front side (d) 8.33 cm from the lens on the front side

Explanation / Answer

The focal length of the lens f = 10 cm (a) When the image distance q = 20 cm since for real image the distance is positive then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (20) / 20 - 10                              = 20 cm from the lens on the front (b) When the image distance q = 50 cm then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (50) / 50 - 10                              = 12.5 cm from the lens on the front (c) When the image distance is q = - 20 cm since the image is virtual     then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-20) / -20 - 10                              = - 200 / -30                                = 6.67 cm from the lens on the front (d) When the image distance is q = -50cm then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-50) / -50 - 10                               = - 500 / -60                              = 8.33 cm from the lens on the front              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (50) / 50 - 10                              = 12.5 cm from the lens on the front (c) When the image distance is q = - 20 cm since the image is virtual     then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-20) / -20 - 10                              = - 200 / -30                                = 6.67 cm from the lens on the front (d) When the image distance is q = -50cm then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-50) / -50 - 10                               = - 500 / -60                              = 8.33 cm from the lens on the front              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-20) / -20 - 10                              = - 200 / -30                                = 6.67 cm from the lens on the front (d) When the image distance is q = -50cm then from lens formla              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-50) / -50 - 10                               = - 500 / -60                              = 8.33 cm from the lens on the front              1/f = 1/p + 1/q then the object distance        p = f q / q -f   = (10) (-50) / -50 - 10                               = - 500 / -60                              = 8.33 cm from the lens on the front

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