Question

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under :

Position of candle =12.0 cm
Position of convex lens =50.0 cm
Position of the screen =88.0 cm
1) where will the image be formed if he shifts the candle towards the lens at a position of 31.0

Answer 1

→Position of the candle flame = 12.0cm

→Position of the lens= 50.0 cm

→Position of the screen=88.0 cm

i) u= 50-12= 38 cm.

→Image distance (v)= 88-50= 38cm

→Focal length =$\large\bf\red{1/v – 1/u= 1/f</p><p>}$

f= 19cm

ii) Object distance u= 50-31= 19 cm

Here,

$\</strong><strong>s</strong><strong>m</strong><strong>a</strong><strong>l</strong><strong>l</strong><strong>\bf\red{</strong><strong>→</strong><strong>Object</strong><strong> </strong><strong>\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong>distance</strong><strong> = focal </strong><strong>\</strong><strong>:</strong><strong>\</strong><strong>:</strong><strong>length</strong><strong>}$

Hence, the image is formed at infinity.

iii) If he further shifts the candle towards the lens. The object comes between F and 0. In this case. Image is virtual, enlarged and erect and is formed on the same side of the lens.

Answer 2

Answer:

umm....

Explanation:

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