physics : class12 Combined Lens Magnification : Combination of Thin Lenses in Contact : Optics MCQ Questions and Answers
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Combined Lens Magnification
MCQ questions and answers.
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MCQ Questions
1. A converging lens of focal length 20 cm and a diverging lens of focal length 10 cm are placed in contact. An object is placed 30 cm in front of the combination. What is the transverse magnification of the final image using the Cartesian sign convention?
A.
+2
B.
-2
C.
+1/2
D.
-1/2
Correct Answer:
+2
Explanation:
The equivalent focal length satisfies 1/F = 1/20 - 1/10 = -1/20, so F = -20 cm. With u = -30 cm, 1/v = 1/F + 1/u = -1/20 + 1/30 = -1/60, giving v = -60 cm. Hence magnification m = v/u = (-60)/(-30) = +2.
2. Two converging lenses of focal lengths 12 cm and 24 cm are placed in contact. An object is positioned 36 cm to the left of the combination. What is the magnification of the final image?
A.
-2/7
B.
+2/7
C.
-7/2
D.
+7/2
Correct Answer:
-2/7
Explanation:
The equivalent focal length is F = (12 × 24)/(12 + 24) = 8 cm. Using 1/v - 1/u = 1/F with u = -36 cm gives 1/v = 1/8 - 1/36 = 7/72, so v = 72/7 cm. Therefore m = v/u = (72/7)/(-36) = -2/7.
3. A combination of two thin lenses in contact forms a real image with magnification -2 when an object is placed 24 cm in front of it. What is the equivalent focal length of the combination?
A.
8 cm
B.
16 cm
C.
24 cm
D.
48 cm
Correct Answer:
16 cm
Explanation:
Since m = v/u, v = (-2)(-24) = 48 cm. Applying 1/F = 1/v - 1/u gives 1/F = 1/48 - 1/(-24) = 1/48 + 1/24 = 1/16. Thus F = 16 cm.
4. A lens combination in contact has an equivalent focal length of 12 cm. For what object distance will its final image be virtual, erect, and three times the size of the object?
A.
-8 cm
B.
-18 cm
C.
-24 cm
D.
+8 cm
Correct Answer:
-8 cm
Explanation:
For an erect image magnified three times, m = +3, so v = 3u. Substituting into 1/v - 1/u = 1/F gives 1/(3u) - 1/u = 1/12, hence -2/(3u) = 1/12. Therefore u = -8 cm, confirming a virtual, erect image.
5. A converging lens of focal length 10 cm and a converging lens of focal length 20 cm are in contact. An object is 30 cm from the combination. If the lenses are treated successively, what is the product of their individual transverse magnifications?
A.
-2/7
B.
+2/7
C.
-7/2
D.
+7/2
Correct Answer:
-2/7
Explanation:
The first lens gives v1 = 15 cm and m1 = 15/(-30) = -1/2. The intermediate image acts as a virtual object for the second lens, giving m2 = 4/7. Therefore the combined magnification is m1m2 = (-1/2)(4/7) = -2/7.
6. Two thin lenses in contact produce an image with magnification +1 when an object is placed 30 cm in front of them. If one lens has focal length 30 cm, what must be the focal length of the other lens?
A.
-30 cm
B.
-20 cm
C.
-15 cm
D.
+15 cm
Correct Answer:
-30 cm
Explanation:
For m = +1, v = u = -30 cm, so 1/F = 1/v - 1/u = 0. Therefore the net power is zero: 1/30 + 1/f2 = 0. Hence f2 = -30 cm.
7. Two converging lenses of focal lengths 15 cm and 30 cm are in contact. An object is placed 45 cm in front of the combination. What is the magnitude and nature of the final magnification?
A.
2/7, real and inverted
B.
7/2, real and inverted
C.
2/7, virtual and erect
D.
7/2, virtual and erect
Correct Answer:
2/7, real and inverted
Explanation:
The equivalent focal length is F = 10 cm. For u = -45 cm, 1/v = 1/10 - 1/45 = 7/90, so v = 90/7 cm. Thus m = v/u = -2/7, indicating a real, inverted image with magnitude 2/7.
8. A lens combination has an equivalent focal length of -15 cm. An object is placed 30 cm in front of it. What is the combined magnification?
A.
+1
B.
-1
C.
+2
D.
-2
Correct Answer:
+1
Explanation:
Using F = -15 cm and u = -30 cm, 1/v = 1/F + 1/u = -1/15 + 1/30 = -1/30, so v = -30 cm. Therefore m = v/u = (-30)/(-30) = +1, representing an erect image of equal size.
9. A lens combination has equivalent focal length 10 cm. For an object at 30 cm, its magnification is -2/7. If the object is moved to 20 cm from the combination, what will be the new magnification?
A.
-1
B.
+1
C.
-2/7
D.
+2/7
Correct Answer:
-1
Explanation:
For u = -20 cm, 1/v = 1/10 - 1/20 = 1/20, so v = 20 cm. Hence the new magnification is m = v/u = 20/(-20) = -1. The change follows directly from the combined-lens imaging relation.
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