JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 4)
To find the focal length of a convex mirror, a student records the following data :
The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :
| Object Pin |
Convex Lens |
Convex Mirror |
Image Pin |
|---|---|---|---|
| 22.2 cm | 32.2 cm | 45.8 cm | 71.2 cm |
The focal length of the convex lens is f1 and that of mirror is f2. Then taking index correction to be negligibly small, f1 and f2 are close to :
f1 = 12.7 cm f2 = 7.8 cm
f1 = 7.8 cm f2 = 12.7 cm
f1 = 7.8 cm f2 = 25.4 cm
f1 = 15.6 cm f2 = 25.4 cm
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For lens :
u1 = $$-$$ (32.2 $$-$$ 22.2) cm
= $$-$$ 10 cm
v1 = (71.2 $$-$$ 32.2) cm
= 39 cm
$$ \therefore $$ $${1 \over {{f_1}}}$$ = $${1 \over {{v_1}}} - {1 \over {{u_1}}}$$
= $${1 \over {39}}$$ + $${1 \over {10}}$$
= $${{49} \over {390}}$$
$$ \therefore $$ f1 = 7.8 cm
For mirror :
R = (71.2 $$-$$ 45.8) cm
= 25.4 cm
$$ \therefore $$ f2 = $${R \over 2}$$ = $${{25.4} \over 2}$$ = 12.7 cm
u1 = $$-$$ (32.2 $$-$$ 22.2) cm
= $$-$$ 10 cm
v1 = (71.2 $$-$$ 32.2) cm
= 39 cm
$$ \therefore $$ $${1 \over {{f_1}}}$$ = $${1 \over {{v_1}}} - {1 \over {{u_1}}}$$
= $${1 \over {39}}$$ + $${1 \over {10}}$$
= $${{49} \over {390}}$$
$$ \therefore $$ f1 = 7.8 cm
For mirror :
R = (71.2 $$-$$ 45.8) cm
= 25.4 cm
$$ \therefore $$ f2 = $${R \over 2}$$ = $${{25.4} \over 2}$$ = 12.7 cm