Question

If two convex lenses of focal lengths
15 cm and 10 cm, separated by a
distance of 8 cm, the focal length of
the combined lens system is
A) 8.00 cm
B) 8.82 cm
C) 7.82 cm
D) 6.82 cm​

Answer 1

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$

Answer:

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$

Explanation:

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$

Answer 2

Explanation:

B) 8.82

I think this the answer...