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Q. No.A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be:
1. increased two times
2. increased times \(\dfrac{3}{2}\) times
3. decreased two times
4. decreased times \(\dfrac{3}{2}\)
1. increased two times
2. increased times \(\dfrac{3}{2}\) times
3. decreased two times
4. decreased times \(\dfrac{3}{2}\)
Subtopic: Simple & Compound Microscope |
Level 3: 35%-60%
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In a microscope the objective is having focal length \(f_0=2~\text{cm}\) and eye-piece is having focal length \(f_e=4~\text{cm}.\) The tube length is \(32~\text{cm}.\) The magnification produced by this microscope for normal adjustment is:
1. \(50\)
2. \(100\)
3. \(150\)
4. \(200\)
1. \(50\)
2. \(100\)
3. \(150\)
4. \(200\)
Subtopic: Simple & Compound Microscope |
85%
Level 1: 80%+
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In a microscope of tube length \(10~\text{cm}\) two convex lenses are arranged with focal length of \(2~\text{cm} \) and \(5~\text{cm}\). Total magnification obtained with this system for normal adjustment is \((5)^k.\) The value of \(k\) is:
1. \(2\)
2. \(5\)
3. \(3.5\)
4. \(4\)
1. \(2\)
2. \(5\)
3. \(3.5\)
4. \(4\)
Subtopic: Simple & Compound Microscope |
79%
Level 2: 60%+
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An object viewed from a near point distance of \(25~\text{cm},\) using a microscopic lens with magnification \('6’,\) gives an unresolved image, a resolved image is observed at an infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length \(0.6~\text{m},\) if the focal length of the eyepiece is equal to:
1. \(15~\text{cm}\)
2. \(20~\text{cm}\)
3. \(22~\text{cm}\)
4. \(25~\text{cm}\)
1. \(15~\text{cm}\)
2. \(20~\text{cm}\)
3. \(22~\text{cm}\)
4. \(25~\text{cm}\)
Subtopic: Simple & Compound Microscope |
56%
Level 3: 35%-60%
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Given below are two statements
| Assertion (A): | For a simple microscope, the angular size of the object equals the angular size of the image. |
| Reason (R): | Magnification is achieved as the small object can be kept much closer to the eye than \(25\) cm and hence it subtends a large angle. |
| 1. | Both (A) and (R) are true and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | (A) is false but (R) is true. |
Subtopic: Simple & Compound Microscope |
Level 3: 35%-60%
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If we need a magnification of \(375\) from a compound microscope of tube length \(150~\text{mm}\) and an objective of focal length \(5~\text{mm},\) the focal length of the eyepiece should be close to:
1. \(22~\text{mm}\)
2. \(33~\text m\)
3. \(12~\text m\)
4. \(2~\text{mm}\)
1. \(22~\text{mm}\)
2. \(33~\text m\)
3. \(12~\text m\)
4. \(2~\text{mm}\)
Subtopic: Simple & Compound Microscope |
Level 3: 35%-60%
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A compound microscope has an objective lens of focal length \(1~\text{cm},\) a tube length of \(20~\text{cm},\) and forms a magnified virtual image \(25~\text{cm}\) from the eyepiece; if its total magnification is \(100,\) the focal length of the eyepiece is:
1. \(3\) cm
2. \(6.25\) cm
3. \(10\) cm
4. \(15\) cm
Subtopic: Simple & Compound Microscope |
79%
Level 2: 60%+
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Q. No.
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