The equation relating the focal length f of the lens to the object distance p 1 .

Question

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Answer 1

Question

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

(c)

To determine

The value of p1 .

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

(d)

To determine

The focal length of the lens.

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

The equation relating the focal length f of the lens to the object distance p1 is,

1f=1p1+11.50 m−p1

Substitute 0.30 m for p1 in above equation.

1f=10.30 m+11.50 m−0.30 mf=0.240 m

Thus, the focal length of the lens is 0.240 m .

Conclusion:

Therefore, the focal length of the lens is 0.240 m .

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Answer 2

Question

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

(c)

To determine

The value of p1 .

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

(d)

To determine

The focal length of the lens.

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

The equation relating the focal length f of the lens to the object distance p1 is,

1f=1p1+11.50 m−p1

Substitute 0.30 m for p1 in above equation.

1f=10.30 m+11.50 m−0.30 mf=0.240 m

Thus, the focal length of the lens is 0.240 m .

Conclusion:

Therefore, the focal length of the lens is 0.240 m .

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Answer 3

Question

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

(c)

To determine

The value of p1 .

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

(d)

To determine

The focal length of the lens.

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

The equation relating the focal length f of the lens to the object distance p1 is,

1f=1p1+11.50 m−p1

Substitute 0.30 m for p1 in above equation.

1f=10.30 m+11.50 m−0.30 mf=0.240 m

Thus, the focal length of the lens is 0.240 m .

Conclusion:

Therefore, the focal length of the lens is 0.240 m .

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Answer 4

Question

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

(c)

To determine

The value of p1 .

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

(d)

To determine

The focal length of the lens.

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

The equation relating the focal length f of the lens to the object distance p1 is,

1f=1p1+11.50 m−p1

Substitute 0.30 m for p1 in above equation.

1f=10.30 m+11.50 m−0.30 mf=0.240 m

Thus, the focal length of the lens is 0.240 m .

Conclusion:

Therefore, the focal length of the lens is 0.240 m .

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Answer 5

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

(c)

To determine

The value of p1 .

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

(d)

To determine

The focal length of the lens.

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

The equation relating the focal length f of the lens to the object distance p1 is,

1f=1p1+11.50 m−p1

Substitute 0.30 m for p1 in above equation.

1f=10.30 m+11.50 m−0.30 mf=0.240 m

Thus, the focal length of the lens is 0.240 m .

Conclusion:

Therefore, the focal length of the lens is 0.240 m .

Answer 6

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Answer 7

Chapter 36, Problem 90AP

(a)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

Answer 8

(a)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p1+1q1 (1)

Here,

q1 is the distance of image from the lens.

The sum of image and object distance is,

p1+q1=1.50 mq1=1.50 m−p1

Substitute 1.50 m−p1 for q1 in equation (1).

1f=1p1+11.50 m−p1 . (2)

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+11.50 m−p1 .

Answer 13

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Answer 14

(b)

To determine

The equation relating the focal length f of the lens to the object distance p1 .

Answer 15

(b)

Expert Solution

Answer to Problem 90AP

The equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Write the expression of lens equation.

1f=1p2+1q2 (3)

Here,

p2 is the final object distance.

q2 is the final image distance.

The final distance of object is,

p2=p1+90 cm×10−2 m1 cm=p1+0.900 m

The final distance of image is,

q2=q1−90 cm×10−2 m1 cm=q1−0.900 m

Substitute 1.50 m−p1 for q1 in above equation.

q2=1.50 m−p1−0.900 m=0.600 m−p1

Substitute 0.600 m−p1 for q2 and p1+0.900 m for p2 in equation (3).

1f=1p1+0.900 m+10.600 m−p1 (4)

Thus, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Conclusion:

Therefore, the equation relating the focal length f of the lens to the object distance p1 is 1f=1p1+0.900 m+10.600 m−p1 .

Answer 20

(c)

To determine

The value of p1 .

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

Answer 21

(c)

To determine

The value of p1 .

Answer 22

(c)

Expert Solution

Answer to Problem 90AP

The value of p1 is 0.30 m .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

Compare equation (2) and (4).

1p1+11.50 m−p1=1p1+0.900 m+10.600 m−p11.50 m−p1+p2p1(1.50 m−p1)=0.600 m−p1+p1+0.900 m(p1+0.900 m)(0.600 m−p1)p1(1.50 m−p1)=(p1+0.900 m)(0.600 m−p1)p1=0.30 m

Conclusion:

Therefore, the value of p1 is 0.30 m .

Answer 27

(d)

To determine

The focal length of the lens.

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .

The equation relating the focal length f of the lens to the object distance p1 is,

1f=1p1+11.50 m−p1

Substitute 0.30 m for p1 in above equation.

1f=10.30 m+11.50 m−0.30 mf=0.240 m

Thus, the focal length of the lens is 0.240 m .

Conclusion:

Therefore, the focal length of the lens is 0.240 m .

Answer 28

(d)

To determine

The focal length of the lens.

Answer 29

(d)

Expert Solution

Answer to Problem 90AP

The focal length of the lens is 0.240 m .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The distance between the candle and the wall is 1.50 m . The distance by which the lens is moved is 90.0 cm . The focal length of the lens is f , the object distance is p1 .