Sponsor Area

Acids, Bases And Salts

Question
CBSEENSC10014500

A student has obtained an image of a distant object on a screen to determine the focal length F1 of the given lens. His teacher, after checking the image, gave him another lens of focal length F2 and asked him to focus the same object on the same screen. The student found that to obtain a sharp image, he has to move the lens away from the screen. From this finding, we may conclude that both the lens given to the students were:

  • Concave and F1 < F2

  • Convex and F1< F2

  • Convex and F1> F2

  • Concave and F1> F2

Solution

C.

Convex and F1> F2

Here, the correct option is C.

Since the image formed is real, the lens used is convex. Given that, the image distance is increasing therefore, the object distance is decreasing. 
Using space the space lens space formula comma

1 over straight F space equals 1 over straight v minus 1 over straight u

straight F equals space fraction numerator uv over denominator straight u minus straight v end fraction

object space distance space equals negative straight u space for space convex space len.
imgae space distance space plus straight v
So comma space Focal space length comma space straight F space equals fraction numerator uv over denominator straight u plus straight v end fraction
Hence comma space straight F space is space less space for space the space second space lens space of space focal space length space straight F subscript 2. space So comma space straight F subscript 1 greater than straight F subscript 2. end subscript