SATHEE: Optics 7 Question 2

Optics 7 Question 2

2. An upright object is placed at a distance of $40 c m$ in front of a convergent lens of focal length $20 c m$ . A convergent mirror of focal length $10 c m$ is placed at a distance of $60 c m$ on the other side of the lens. The position and size of the final image will be

(2019 Main, 8 April I)

(a) $20 c m$ from the convergent mirror, same size as the object

(b) $40 c m$ from the convergent mirror, same size as the object

(d) $20 c m$ from the convergent mirror,twice size of the object

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Solution:

In given system of lens and mirror, position of object $O$ in front of lens is at a distance $2 f$ .

i.e. $u = 2 f = 40 c m$

So, image $(I_{1})$ formed is real, inverted and at a distance, $v = 2 f = 2 \times 20 = 40 c m$ , (behind lens) magnification, $m_{1} = \frac{v}{u} = \frac{40}{40} = 1$

Thus, size of image is same as that of object.

This image $(I_{1})$ acts like a real object for mirror.

As object distance for mirror is $u = C = 2 f = - 20 c m$ where, $C =$ centre of curvature.

So, image $(I_{2})$ formed by mirror is at $2 f$ .

$∴$ For mirror $v = 2 f = 2 (- 10) = - 20 c m$

Magnification, $m_{2} = - \frac{v}{u} = - \frac{(- 20)}{(- 20)} = - 1$

Thus, image size is same as that of object.

The image $I_{2}$ formed by the mirror will act like an object for lens.

As the object is at $2 f$ distance from lens, so image $(I_{3})$ will be formed at a distance $2 f$ or $40 c m$ . Thus, magnification,

$m_{3} = \frac{v}{u} = \frac{40}{40} = 1$

So, final magnification, $m = m_{1} \times m_{2} \times m_{3} = - 1$

Hence, final image $(I_{3})$ is real, inverted of same size as that of object and coinciding with object.

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Optics 7 Question 2

2. An upright object is placed at a distance of $40 c m$ in front of a convergent lens of focal length $20 c m$ . A convergent mirror of focal length $10 c m$ is placed at a distance of $60 c m$ on the other side of the lens. The position and size of the final image will be

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics 7 Question 2

2. An upright object is placed at a distance of 40cm in front of a convergent lens of focal length 20cm. A convergent mirror of focal length 10cm is placed at a distance of 60cm on the other side of the lens. The position and size of the final image will be

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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2. An upright object is placed at a distance of $40 c m$ in front of a convergent lens of focal length $20 c m$ . A convergent mirror of focal length $10 c m$ is placed at a distance of $60 c m$ on the other side of the lens. The position and size of the final image will be