The double concave lens in the figure above isfabricated of glass having index o

Question

The double concave lens in the figure above isfabricated of glass having index of refraction n = 1.75.The magnitudes of the radii of curvature of theleft and right surfaces of the lens are |Rl| =22 cm and |Rr| = 17 cm, respectively.

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

P = -12.7863247863248 diopters    

An upright object of height H = 0.7 cm is placed at adistance of 19 cm to the left of the lens.

(b) Calculate the image distance.

s1' = cm    

(c) Calculate 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 =     

The object is removed and instead converging rays are incomingfrom the left. In the absence of the lens, these rays would form anupright image at a position 6 cm to the right ofthe location of the lens, i.e. a virtualobject.

(f) Calculate the image distance in this case. (Give numericalvalue and proper 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 =     

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

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

Explanation / Answer

Given :                  n      = 1.75                  R1   = 22 cm                  R2   = 17 cm For concave lens :                1 / f    =   (n -1 ) [ 1 /R1    - 1 / R2 ]                          =   (1.75 - 1) [ 1 / 22 - 1/ 17 ]                Focal length (f)   = - 99.73 cm Power is :                    P   = 1 / f   =   - 1 /99.73                                      =    - 0.01 (b)           Object distance is (u)   = 19 cm We have :              1 / f    = 1 / u   +   1 /v         or   1 /v   = 1 / f   -   1 / u                        = - 1 / 99.73   - 1 / 19           Imagedistance is :                   v   =   -15.96  cm (c)             We have :               Magnification (M) = - v / u   = hi / ho                             or        15.96 / 19   =   hi / 0.7                                               hi      =   (15.96 * 0.7) / 19                                                             = 0.588 cm (d)               Imag is virtual (e)              Image is upright. (f)           New Image distance is :                    S2    = -15.96 - 6 cm   = -21.96   cm (g)           Imageis real              Inverted           Locatedleft to the lens. Hope this helps u!     

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