Question

(3) 24°
A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If this
telescope is used to see a 50 m tall building at a distance of 2 km, what is the height of the image of the
building formed by the objective lens?
(2) 10 cm
(1) 5 cm
(4) 2 cm
(3) 1 cm
​

Answer 1

Answer:

option(1): 5 cm

Explanation:

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Answer 2

a$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Focal length of objective lens (f) = 200 cm ⇒ 2 m..
• Focal length of eyepiece (f') = 2 cm ⇒ 0.02 m. [This value is not required because image first forms in the objective lens]
• Distance of object (u) = –2km ⇒ –2000 m (u is always negative)
• Height of object (h) = 50 m.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• Height of image formed by objective lens.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

We will use the lens formula:

$\mapsto \sf{\green{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u}}}$

Where:

• f is focal length.
• v is image distance.
• u is object distance.

Substitute the values.

$\sf \dfrac{1}{2} =\dfrac{1}{v} -\dfrac{1}{-2000}$

$\sf \dfrac{1}{2}-\dfrac{1}{2000} =\dfrac{1}{v}$

$\sf v \approx 2 m.$

Now, we know that magnification:

$\mapsto \sf{\red{\dfrac{h'}{h} =\dfrac{v}{u} }}$

Where:

• h' is image height.
• h is object height.
• v is image distance.
• u is object distance.

Substitute the values.

$\to \sf \dfrac{h'}{50} =\dfrac{2}{-2000}$

$\to \sf h'=\dfrac{1}{20} m.$

$\to \sf h'=\left(\dfrac{1}{20} \times 100 \right) cm.$

$\boxed{\sf{\green{h'=5 cm.}}}$

HEIGHT OF IMAGE IS 5 cm.

∴ 2nd option is correct ✓✓✓.