Question

3&4, thanks!

Question 3 1 pts What is the object distance in the First Camera Image question in terms of X? Question 4 2 pts What is the image vergence in the First Camera Image question in terms of X? For full credit, use Math Equation () to show not just the answer but how you got the answer. HTML Editora BIU A А IX E311 x X, SE V 60 V 1:12pt paragraph

Answer 1

and - 1 / 1 P f =t& length of lens (the sign lens) Q3 given for convering lens IPS L Object vergere u (where negative sign implies that Object is placed loft to the lens) The Power of the objective lens is given by ㅗ where f is focal length of leng So focal is fat for converging The vergence is define by vergence where n = reßactive index of medica el la distance of object or image from lens The sign of vergence is depends on sign convention take incident direction as positive The distance of object from lens Here medium from object vergence the do =+56 de using lens formula I +1 where do de dos distance of the image -th 50 do air So n = ) object vergere, U = ten from lens to tok di do do f do-f (5x1x 58-se 58² 4K 1.25

will So distance of image from lens tens ½ dp = + 1.25* (where the sign of de implies that image from Opbosit side placed ) converging lens as object Object fkk- do moge di- image to (a 1.25x Q4 The vergence is given by (n=l for air) V h de the sign of vergence imply that V= i as object is place left with negative vergence The image will form right to the rens so The image vergence is given by 11. 1.25 x V YTTER