How can we proof that Focal length is half of centre of curvature?
Answers
Proving the focal length is half the radius of curvature:
Taking a concave mirror, the curved mirror will have a principal axis near which a ray of light is incident on the mirror parellel to it. ... Hence, in both cases Radius is double the focal length.
Answer:
1 \frac{3}{4} \: + 4\frac{5}{7} \: + 7\frac{9}{7} \: + 12\frac{6}{25}1
4
3
+4
7
5
+7
7
9
+12
25
6
1 \frac{3}{4} = (\frac{4 \times 1 + 3 }{4} ) \: = \: \frac{7}{4}1
4
3
=(
4
4×1+3
)=
4
7
4 \frac{5}{7} = (\frac{7 \times 4 + 5}{7} ) \: = \frac{33}{7}4
7
5
=(
7
7×4+5
)=
7
33
7 \frac{9}{7} =( \frac{7 \times 7 + 9}{7} )= \frac{58}{7}7
7
9
=(
7
7×7+9
)=
7
58
12 \frac{6}{25} = (\frac{25 \times 12 + 6}{25} )= \frac{306}{25}12
25
6
=(
25
25×12+6
)=
25
306
Now, all are improper fractions convert them into proper fractions, to do that you should do LCM.
Take LCM of 4 ,7, 7, 25
LCM = 2 \times 2 \times 5 \times 5 \times 7 = 700LCM=2×2×5×5×7=700
Now, make them into proper fractions.
\frac{7}{4} \times \frac{175}{175} = \frac{1225}{700}
4
7
×
175
175
=
700
1225
\frac{33}{7 } \times \frac{100}{100} = \frac{3300}{700}
7
33
×
100
100
=
700
3300
\frac{58}{7} \times \frac{100}{100}= \frac{5800}{700}
7
58
×
100
100
=
700
5800
\frac{306}{25} \times \frac{28}{28}= \frac{8568}{700}
25
306
×
28
28
=
700
8568
Now, add all the proper fractions
( \frac{1225 + 3300 + 5800 + 8568}{700} )(
700
1225+3300+5800+8568
)
\frac{18893}{700}
700
18893
26 \frac{693}{700}26
700
693
Hope it helps you