Question
In a compound microscope, the focal length of the objective and the eye lens are 2.5 cm and 5 cm respectively. An object is placed at 3.75 cm before the objective and image is formed at the least distance of distinct vision, then the distance between two lenses will be (i.e. length of the microscopic tube)
Answer: Option A
:
A
LD=vo+ue and for objective lens 1fo=1vo−1uo
Putting the values with proper sigh convention.
1+2.5=1vo−1(−3.57)⇒vo=7.5cm
For eye lens 1fe=1ve−1ue
⇒1+5=1(−25)−1ue⇒ue=−4.16cm
⇒|ue|=4.16cm
Hence LD=7.5+4.16=11.67cm
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:
A
LD=vo+ue and for objective lens 1fo=1vo−1uo
Putting the values with proper sigh convention.
1+2.5=1vo−1(−3.57)⇒vo=7.5cm
For eye lens 1fe=1ve−1ue
⇒1+5=1(−25)−1ue⇒ue=−4.16cm
⇒|ue|=4.16cm
Hence LD=7.5+4.16=11.67cm
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