The double convex lens in the figure above isfabricated of glass with index of r

Question

The double convex lens in the figure above isfabricated of glass with index of refraction n = 1.77. Themagnitudes of the radii of curvature of the leftand right surfaces of the lens are |Rl| = 18 cmand |Rr| = 13 cm, respectively.

(a) Calculate the power P of the lens in"diopters".

P = diopters    

Calculate the image distance. (Give a numerical value and properalgebraic sign.)

s1' = cm    

(c) Calculate the magnitude of the image height.

H' = cm    

(d) Is the image real or virtual?

Answer as follows: Real Image = 1; Virtual Image = 2.

Image type =     

(e) Is the image upright or inverted?

Answer as follows: Upright = 11, Inverted = 22.

Image orientation =     

Calculate the new image distance. (Give numerical value andproper algebraic sign.)

s2' = cm    

(g) Is the image real or virtual?

Answer as follows: Real Image = 1; Virtual Image = 2.

Image type =     

(h) Is the image upright or inverted?

Answer as follows: Upright = 11, Inverted = 22.

Image orientation =     

Calculate the image distance in this case. (Give numerical valueand proper algebraic sign.)

s3' = cm    

(j) Is the image real or virtual?

Answer as follows: Real Image = 1; Virtual Image = 2.

Image type =     

(k) Is the image upright or inverted?

Answer as follows: Upright = 11, Inverted = 22.

Image orientation =     

(l) Is the image located to the right or left of the lens?

Answer as follows: Right = 111, Left = 222.

Explanation / Answer

(a)               We know that :                 Focal length is :                               1 / f    = (n - 1) [ 1/R1   - 1 / R2 ]                                         =   ( 1.77 - 1) [ 1 / 18 - 1 / 13]                      Focal length (f) = - 60.77  cm            Power is :                          P   = 1 / f    =   - 1 /60.77                                             = - 0.0164 dioptors (b)             Given Object distance (u) = 23 cm           We knowthat :                        1 / f    =   1/ u + 1 /v             or       1 / v   =   1 / f - 1 /u                                  =   - 1 / 60.77   - 1 / 23                 Image distance (v)   = -16.68   cm (c)           Magnification (M) = - v / u                                        = 16.68 / 23                                        = 0.7252 (d)           Imageis real (e)           Image is Inverted. (f)           We knowthat :                         1 / f   = 1/ u + 1/ v                or      1 / v = 1 /f   - 1 / u                                   =  - 1 / 60.77 - 1 / (8+ 23)               or     v       = - 20.53  cm (g)           Imageis real and inverted. Similarly do the last one. Hope this helps u!           

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