The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.

Q: The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.

The lens is a **convex lens**. The focal length of the lens is $+30\text{ cm}$ (or $30\text{ cm}$). The height of the image is $-4\text{ cm}$ (the negative sign indicates that the image is real and inverted).

Light — Reflection and Refraction

Refraction by Spherical Lenses

Official Answer

The lens is a convex lens.

The focal length of the lens is $+30\text{ cm}$ (or $30\text{ cm}$ ).

The height of the image is $-4\text{ cm}$ (the negative sign indicates that the image is real and inverted).

convex lensreal imagelens formulafocal length 30 cmmagnificationheight of image -4 cm

Marking Scheme

11.0 mark for identifying the lens as a convex lens with justification.
21.0 mark for correctly calculating the focal length ( $f = +30\text{ cm}$ ) using the lens formula.
31.0 mark for correctly calculating the height of the image ( $h' = -4\text{ cm}$ ) using the magnification formula.

Hint

Since the image is caught on a screen, it must be real. Use the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with proper sign conventions, and then use the magnification formula $m = \frac{h'}{h} = \frac{v}{u}$ .

Quick Oral Answer

A convex lens is used because only it can form a real image on a screen. The focal length is calculated using the lens formula to be $30\text{ cm}$ , and the image height is $-4\text{ cm}$ , indicating a real, inverted, and magnified image.

Analysis & Explanation

A real image is always inverted, which is mathematically confirmed by the negative sign of the image height ( $h' = -4\text{ cm}$ ). Since $|v| > |u|$ ( $90\text{ cm} > 45\text{ cm}$ ), the image is magnified, which is also confirmed by the magnitude of the height ( $4\text{ cm} > 2\text{ cm}$ ).

Common Mistakes

1Using the mirror formula instead of the lens formula (i.e., using $+$ instead of $-$ ).
2Forgetting to apply the sign convention for object distance ( $u = -45\text{ cm}$ ).
3Using $m = -\frac{v}{u}$ (which is for mirrors) instead of $m = \frac{v}{u}$ (for lenses).

Interesting Facts

When an object is placed between $F$ and $2F$ of a convex lens, its real, inverted, and magnified image is formed beyond $2F$ on the other side. Here, $f = 30\text{ cm}$ , so $2f = 60\text{ cm}$ . The object at $45\text{ cm}$ is indeed between $F$ and $2F$ , and the image at $90\text{ cm}$ is beyond $2F$ ( $60\text{ cm}$ ), perfectly matching theory!

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2012?

This question carries 3 marks in the CBSE Class 10 Science 2012 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Light — Reflection and Refraction" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Refraction by Spherical Lenses" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2012 paper?

This is a Short Answer question from Section A in the CBSE Class 10 Science 2012 paper.