सूत्र - `epsilon = phi l`
पहली स्थिति में, `epsilon_(1)+epsilon_(2)=phi xx 8`
तथा दूसरी स्थिति में,
`epsilon_(1)-epsilon_(2)=phi xx 2`
`rArr " " (epsilon_(1)+epsilon_(2))/(epsilon_(1)-epsilon_(2))=(phi xx 8)/(phi xx 2) = 4`
`therefore " " 4epsilon_(1)-4epsilon_(2)=epsilon_(1)+epsilon_(2)`
या `4epsilon_(1)-epsilon_(1)=4epsilon_(2)+epsilon_(2)`
`3epsilon_(1)=5epsilon_(2)`
`(epsilon_(1))/(epsilon_(2))=(5)/(3)`.