A diverging lens ( f = -11 cm) islocated 20.0 cm to the left of a converging len

Question

A diverging lens (f = -11 cm) islocated 20.0 cm to the left of a converging lens (f =30.0 cm). A 3.00-cm-tall object stands tothe left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final imagerelative to the converging lens.
b) What is the height of the final image (including properalgebraic sign)?
I know I must use the thins lens equation twice, but whichlens do I calculate first? Also how do you know the objectdistance?Thanks (a) Determine the distance of the final imagerelative to the converging lens.
b) What is the height of the final image (including properalgebraic sign)?
I know I must use the thins lens equation twice, but whichlens do I calculate first? Also how do you know the objectdistance?Thanks

Explanation / Answer

A ) f1 = focallength of diverging lens = -11 cm f2 =  focallength of conerging lens = 30cm Now the distance between the two lenses is 20.0 cm Now object is located 11 cm to the left of diverging lens do =11 cm ( for diverging lens ) Then apply lens equation for diverging lens 1/f 1 = 1/do + 1/di 1/di = 1/do - 1/f1 = 1/-11 cm   -1/11 cm = -2/11 cm = -0.1818 cm -1 di =- 5.5 cm Thus first image is 5.5 cm to the left of diverging lens thenthis image acts as object to the converging lens Then distance ofobject to the converging lens is do' = 5.5 cm + 20.0 = -25.5cm Then 1/f2 = 1/do' + 1/di' Then 1/di' = 1/f2 - 1/do' = 1/30 - 1/25.5   = -0.00588 cm-1 Then image distance from converginglens is di ' =-169.99 cm B) Now we know that magnification m = hi/ho =-di/do ho = 3.0 cm Then height of first image is hi = -(di/do)ho =-(-5.5 cm/11 cm ) * 3 cm                                                                  = 1.5 cm Thus height of object for converginglens is ho' = hi = 1.5cm Then imageis object to converging lens cm Then hi'/ho' = -di'/do'           hi' =-ho'(di'/do') = - 1.5 cm'(-169.99cm/25.5 cm) = 9.99cm di =- 5.5 cm Thus first image is 5.5 cm to the left of diverging lens thenthis image acts as object to the converging lens Then distance ofobject to the converging lens is do' = 5.5 cm + 20.0 = -25.5cm Then 1/f2 = 1/do' + 1/di' Then 1/di' = 1/f2 - 1/do' = 1/30 - 1/25.5   = -0.00588 cm-1 Then image distance from converginglens is di ' =-169.99 cm B) Now we know that magnification m = hi/ho =-di/do ho = 3.0 cm Then height of first image is hi = -(di/do)ho =-(-5.5 cm/11 cm ) * 3 cm                                                                  = 1.5 cm Thus height of object for converginglens is ho' = hi = 1.5cm Then imageis object to converging lens cm Then hi'/ho' = -di'/do'           hi' =-ho'(di'/do') = - 1.5 cm'(-169.99cm/25.5 cm) = 9.99cm

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