A transparent photographic slide is placed in front of a converging lens with a

Question

A transparent photographic slide is placed in front of a converging lens with a focal length of 2.48 cm. The lens forms an image of the slide 12.5 cm from the slide. (a) How far is the lens from the slide if the image is real? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = cm

further possible distance = cm
(b) How far is the lens from the slide if the image is virtual? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = cm further possible distance = cm




A transparent photographic slide is placed in front of a converging lens with a focal length of 2.48 cm. The lens forms an image of the slide 12.5 cm from the slide. (a) How far is the lens from the slide if the image is real? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = cm

further possible distance = cm
(b) How far is the lens from the slide if the image is virtual? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = cm further possible distance = cm



(a) How far is the lens from the slide if the image is real? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = cm

further possible distance = cm
(b) How far is the lens from the slide if the image is virtual? (Enter 'none' in the second box if there is only one answer.) nearer possible distance = cm further possible distance = cm nearer possible distance = cm

further possible distance = cm

Explanation / Answer

a) 1/f = 1/u + 1/v

1/2.48 – 1/12.5 = 1/u

u = 3.09 cm

b) 1/u = 1/f + 1/v ( in this case v is negative and so becomes positive when taken to the other side of the equation )

u = 2.06 cm

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