JEE MAIN - Physics (2021 - 27th August Morning Shift - No. 14)
Find the distance of the image from object O, formed by the combination of lenses in the figure :
_27th_August_Morning_Shift_en_14_1.png)
75 cm
10 cm
20 cm
infinity
Explanation
$${1 \over {{V_1}}} + {1 \over {30}} = {1 \over {10}}$$
$${1 \over {{V_1}}} = {2 \over {30}} \Rightarrow {V_1} = 15$$ cm
$${1 \over {{V_2}}} - {1 \over {10}} = - {1 \over {10}}$$
$${1 \over {{V_2}}} = 0$$
V2 = $$\infty$$
V3 = 30 cm
OV3 = 75 cm
$${1 \over {{V_1}}} = {2 \over {30}} \Rightarrow {V_1} = 15$$ cm
$${1 \over {{V_2}}} - {1 \over {10}} = - {1 \over {10}}$$
$${1 \over {{V_2}}} = 0$$
V2 = $$\infty$$
V3 = 30 cm
OV3 = 75 cm
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