Please help

An image of small size can be obtained on a screen placed at 2.0 m from object. It can be done by using:

(A) concave mirror of focal length 0.5 m
(B) concave mirror of focal length greater than 0.5 m
(C) concave mirror of focal length less than 0.5 m
(D) concave lens of focal length less than 0.5 m.

Can you provide a good solution for this? Thank you

#Light

Easy Math Editor

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Comments

What have you tried? What do you know?

Calvin Lin Staff - 5 years, 11 months ago

I know that option D is not true because a concave lens won't produce an image which can be obtained on a screen, i.e it won't produce a real image. The things which i can't figure out are the three options left. In the question the object distance is not given. 2.0 m is mere distance of screen from the object, i.e. image from the object.

How can i figure out the focal length of the mirror?

Sahba Hasan - 5 years, 11 months ago

@Calvin Lin , @Nihar Mahajan , @Jake Lai ... Please help...

Sahba Hasan - 5 years, 10 months ago

B will be the option as in that case, the object will be placed between centre of curvature and focal length. Since the object is placed at 2 m away from the mirror, and it has to be in between Curvature and focus. The centre of curvature therefore should be greater than 2 m which makes focal length automatically greater than 0.5 m.

Abhay Tiwari - 5 years, 11 months ago

You got it wrong @Abhay Tiwari ...The object distance is not given, the distance of the image from the object is given.

Sahba Hasan - 5 years, 10 months ago

Still the answer remains B :p

Abhay Tiwari - 5 years, 10 months ago

@Abhay Tiwari – I can't understand how...@Abhay Tiwari ...

Sahba Hasan - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$