12th Grade > Physics
OPTICAL INSTRUMENTS AND SOME NATURAL PHENOMENA MCQs
Total Questions : 30
| Page 2 of 3 pages
Answer: Option B. -> Smaller focal length
:
B
m=1+D/f.So, smaller the focal length, higher the magnifying power.
:
B
m=1+D/f.So, smaller the focal length, higher the magnifying power.
Answer: Option C. -> −14D
:
C
By using f=xyx−y⇒f=3×123−12=−4m. Hence power P=1f=−14D
:
C
By using f=xyx−y⇒f=3×123−12=−4m. Hence power P=1f=−14D
Question 14. A compound microscope is used to enlarge an object kept at a distance 0.03m from it’s objective which consists of several convex lenses in contact and has focal length 0.02m. If a lens of focal length 0.1m is removed from the objective, then by what distance the eye-piece of the microscope must be moved to refocus the image
Answer: Option D. -> 9 cm
:
D
If initially the objective (focal length F0) forms the image at distance v0then v0=u0f0u0−f0=3×23−2=6cm
Now as in case of lenses in contact 1F0=1f1+1f2+1f3+.....=1f1+1F′0{where1F′0=1f2+1f3+...}
So if one of the lens is removed, the focal length of the remaining lens system
1F′0=1F0−1f1=12−110⇒F′0=2.5cm
This lens will form the image of same object at a distance v′0such that v′0=u0F′0u0−F′0=3×2.5(3−2.5)=15cm
So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15 – 6 = 9 cm.
:
D
If initially the objective (focal length F0) forms the image at distance v0then v0=u0f0u0−f0=3×23−2=6cm
Now as in case of lenses in contact 1F0=1f1+1f2+1f3+.....=1f1+1F′0{where1F′0=1f2+1f3+...}
So if one of the lens is removed, the focal length of the remaining lens system
1F′0=1F0−1f1=12−110⇒F′0=2.5cm
This lens will form the image of same object at a distance v′0such that v′0=u0F′0u0−F′0=3×2.5(3−2.5)=15cm
So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted i.e. 15 – 6 = 9 cm.
Answer: Option D. -> –1.0 D
:
D
Hey, like you just learned, for correcting the myopic eye, the concave corrective lens should form the image for an object at infinity at the far point of the defective eye.
Applying sign convention, f = –(far point of myopic eye) = – 100 cm. So power of the lens P=100f=100−100=−1D
:
D
Hey, like you just learned, for correcting the myopic eye, the concave corrective lens should form the image for an object at infinity at the far point of the defective eye.
Applying sign convention, f = –(far point of myopic eye) = – 100 cm. So power of the lens P=100f=100−100=−1D
Answer: Option A. -> 0.25 m
:
A
Remember this is the distance nearer to which a normal eye can't see properly, on average.
:
A
Remember this is the distance nearer to which a normal eye can't see properly, on average.
Answer: Option B. -> 1+Df
:
B
This is the case when image is formed at near point.
:
B
This is the case when image is formed at near point.
Answer: Option D. -> 44 D to 40 D
:
D
An eye sees distant objects with full relaxation so 12.5×10−2−1−∞=1f or
P=1f=125×10−2=40D
An eye sees an object at 25 cm with strain so 12.5×10−2−1−25×10−2=1forP=1f=40+4=44D
:
D
An eye sees distant objects with full relaxation so 12.5×10−2−1−∞=1f or
P=1f=125×10−2=40D
An eye sees an object at 25 cm with strain so 12.5×10−2−1−25×10−2=1forP=1f=40+4=44D
Answer: Option A. -> + 3.0 D
:
A
P=1f=1v−1u.
Applying sign convention,
P=1f=1−100−1−25=3100=+3D.
:
A
P=1f=1v−1u.
Applying sign convention,
P=1f=1−100−1−25=3100=+3D.
Answer: Option D. -> 3256cm
:
D
The length of the telescope, LD=f0+ue, as the objective forms the image at its focus, as the rays are from infinity, and this image acts as the object for the eyepiece.
Now, using lens formula, and using the fact that the eyepiece forms image at near point, that is v=D(=−25cm):
1f=1v−1u.
Applying sign convention,1ue=−1D−1fe.
Or, considering only the modulus, ue=feDfe+D.
Therefore,LD=f0+ue=f0+feDfe+D=50+5×25(5+25)=3256cm.
:
D
The length of the telescope, LD=f0+ue, as the objective forms the image at its focus, as the rays are from infinity, and this image acts as the object for the eyepiece.
Now, using lens formula, and using the fact that the eyepiece forms image at near point, that is v=D(=−25cm):
1f=1v−1u.
Applying sign convention,1ue=−1D−1fe.
Or, considering only the modulus, ue=feDfe+D.
Therefore,LD=f0+ue=f0+feDfe+D=50+5×25(5+25)=3256cm.
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