A telescope is said to be in normal adjustment when the
focal length of he objective is greater than that of the eyepiece
focal length of the eyepiece is greater than that of the objective
focal length of the eyepiece is equal to that of the objective
objective focal point coincides with that of the eyepiece
Correct answer is D
No explanation has been provided for this answer.
3.0
1.5
0.6
0.3
Correct answer is C
* produces a virtual image when object distance(u) is less than its focal length(f).
\(\frac{1}{f} = \frac{1}{u} - \frac{1}{v}\)
\(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)
\(\frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30}\)
\(\frac{1}{v} = \frac{5}{30}\)
v = 6.0 cm
but m = \(\frac{v}{u} = \frac{6}{10} = 0.6\)
Note that the negative sign only shows the placement of the image. It has no effect on the magnification basically.
F and X
X and C
C and F
F and 2F
Correct answer is B
No explanation has been provided for this answer.
0.50
0.67
0.75
1.13
Correct answer is D
\(_{g} \eta _{w} = \frac{ _g \eta _a}{ _a \eta _w}\)
= \(\frac{3}{2} \times \frac{1}{\frac{4}{3}}\)
= \(\frac{3}{2} \times \frac{3}{4}\)
= \(\frac{9}{8}\)
= 1.125
= 1.13 (to 2 d.p)
2.5 cm
3.0cm
5.0cm
25.0cm
Correct answer is A
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
= \(\frac{1}{10} - \frac{1}{2} = \frac{1}{20}\)
v = 20
but h\(_i\) = \(\frac{v \times h_o}{u}\)
= \(\frac{20 \times 2.5 }{20}\)
= 2.5cm