please show me step by step how to answer #6and #7 4. [1pt] Suppose you have a c

Question

please show me step by step how to answer #6and #7
4. [1pt]
Suppose you have a converging lens with focal length 25.0 cm andyou place an object 41.67 cm from the lens. What is the location ofthe image?
62.5cm
5. [1pt]
What is the magnification of the image?
-1.5

6. [1pt]
If a diverging lens of focal length -76.0 cm is then placed 25.0 cmfrom the first lens on the opposite side of the lens from theobject, what is the location of the new image?


7. [1pt]
What is the magnification due to the two-lens system?

please show me step by step how to answer #6 and#7
please show me step by step how to answer #6and #7
4. [1pt]
Suppose you have a converging lens with focal length 25.0 cm andyou place an object 41.67 cm from the lens. What is the location ofthe image?
62.5cm
5. [1pt]
What is the magnification of the image?
-1.5

6. [1pt]
If a diverging lens of focal length -76.0 cm is then placed 25.0 cmfrom the first lens on the opposite side of the lens from theobject, what is the location of the new image?


7. [1pt]
What is the magnification due to the two-lens system?

please show me step by step how to answer #6 and#7
please show me step by step how to answer #6and #7
4. [1pt]
Suppose you have a converging lens with focal length 25.0 cm andyou place an object 41.67 cm from the lens. What is the location ofthe image?
62.5cm
5. [1pt]
What is the magnification of the image?
-1.5

6. [1pt]
If a diverging lens of focal length -76.0 cm is then placed 25.0 cmfrom the first lens on the opposite side of the lens from theobject, what is the location of the new image?


7. [1pt]
What is the magnification due to the two-lens system?

please show me step by step how to answer #6 and#7

Explanation / Answer

Focal length of the converging lens = 25.0cm Object distance to the converging lens = 41.67cm Then the image distance of the object to the converging lens =62.5cm Magnification of the converging lens = 1.5 Now if we placed a concave lens to the right of the converginglens, then the image formed due to the converging lens acts as anobject for the diverging lens. It is given that the diverging lens is placed 25.0 cm from thefirst lens on the opposite side of the lens, then the objectdistance to the diverging lens = 62.5cm - 25.0cm                                            = 37.5cm Focal length of the diverging lens = -76.0cm Then (1 / f) = (1 / do) + (1 / di)          -1 /76cm = (1 / 37.5cm) + (1 / di)             di = -25.1cm Now the magnification is m = (1.7)(-25.1cm / 37.5cm)                                          = -1.14 Hope this it may helps you.
Hope this may helps you

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