A 10 cm tall candle, when viewed using a convex lens appears to be 20 cm tall. The focal length is 5 cm. What is the object distance?

10 cm

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5 cm

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20 cm

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2.5 cm

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Solution

The correct option is D
2.5 cm

From len's formula, $\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ .

And, $\frac{v}{u}$ = $\frac{h^{/}}{h}$ = $\frac{20}{10}$

Therefore, $\frac{1}{2 u}$ - $\frac{1}{u}$ = $\frac{1}{5}$

or u = - 2.5 cm

2.5 cm is the object distance.

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Similar questions

A $10 c m$ tall candle when viewed using a convex lens appears to be $30 c m$ tall as shown in the below figure. The focal length of the lens is $5 c m$ . The magnitude of distance of the object is _________ $c m$ .

Q. A

10 c m

tall candle, when viewed using a biconvex lens, appears to be

20 c m

tall. The focal length of the lens is

5 c m

. What is the object distance?

A $10 c m$ tall candle when viewed using a convex lens appears to be $20 c m$ tall as shown in the below figure. The focal length of the lens is $5 c m$ . What is the object distance?

Q. To obtain a magnification of, –2 with a convex lens of focal length 10 cm, the object should be placed:
(a) between 5 cm and 10 cm
(b) between 10 cm and 20 cm
(c) at 20 cm
(d) beyond 20 cm

Q. A

20 c m

tall object is placed perpendicular to the principal axis of a convex lens of focal length

10 c m

. If object distance is

15 c m

. Find the image distance.

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A 10 cm tall candle, when viewed using a convex lens appears to be 20 cm tall. The focal length is 5 cm. What is the object distance?

The correct option is D 2.5 cm From len's formula, 1v - 1u = 1f. And, vu = h/h = 2010 Therefore, 12u - 1u = 15 or u = - 2.5 cm 2.5 cm is the object distance.

The correct option is D
2.5 cm

From len's formula, $\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ .

And, $\frac{v}{u}$ = $\frac{h^{/}}{h}$ = $\frac{20}{10}$

Therefore, $\frac{1}{2 u}$ - $\frac{1}{u}$ = $\frac{1}{5}$

or u = - 2.5 cm

2.5 cm is the object distance.