Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 7 (Q.No. 35)
35.
In a telescope the object glass of focal length 14 cm, is located at 20 cm from the diaphragm. The focussing lens is midway between them when a staff 16.50 m away is focussed. The focal length of the focussing lens, is
Discussion:
9 comments Page 1 of 1.
Annika said:
5 years ago
Let me explain, also tell me if I'm wrong?
1/F1=((1/14)+(1/20))=1/.1214 = 8.237.
1/F2=,((1/16.5)+(1/10))=1/1606 = 6.226.
Note- 10 is taken because its in midway of 20.
Now take the average for F=(f1+f2)/2.
= 7.24.
1/F1=((1/14)+(1/20))=1/.1214 = 8.237.
1/F2=,((1/16.5)+(1/10))=1/1606 = 6.226.
Note- 10 is taken because its in midway of 20.
Now take the average for F=(f1+f2)/2.
= 7.24.
(13)
Deepak Baral said:
5 years ago
Can anyone explain it clearly?
(1)
Aiswarya said:
5 years ago
Can anyone please explain this one
Valavan said:
5 years ago
Can anyone explain this.
Ak singh said:
5 years ago
Please explain it in clear.
(2)
Mohit singh rana said:
7 years ago
I/f = 1/ f1 + 1/f2.
= 1/14 +1/16.50.
1/f = 0.314.
F = 7.24.
= 1/14 +1/16.50.
1/f = 0.314.
F = 7.24.
(1)
Harsha said:
8 years ago
Can anyone explain it clearly?
Gagan said:
8 years ago
When we take the reading by telescope mainly focal length 1, diap = 1.42, the distance will be cover 0.825. Now we have 14 : 20: 16.50.
Neha borade said:
8 years ago
Can anybody explain this?
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