How can we proof that Focal length is half of centre of curvature?
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
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