Answer to Question #282131 in Physics for Ella

Question

3. All real images of a concave mirror are ________

A . Inverted

B. Smaller than the object

C. Behind the minor

D. All of the above

4. The image of an object in front of a convex miror is ________

A. Reduced

B. Also in front of the mirror

C. Inverted

D. Enlarged

5. A penicil 10 cm long is placed 30 cm in front of a mirror of focal length -50 cm. The image is _________

A. 25 cm long and upright

B. 6.3 cm long and upright

C. 25 cm long and inverted

D. 6.3 cm long and inverted

6. An object closer to a converging lens than its focal point always has an image that in ________

A. Inerted

B. Virtual

C. The same in size

D. smaller in size

7. An object is located 10 cm from a converging lens of focal length 12 cm. The image distance is ______

A. +5.5 cm

B. - 5.5cm

C. +60 om

D. -60 cm

Answer 1

3) All real images of a concave mirror are inverted. Therefore, the correct answer is (A).

Answer:

A. Inverted

4) The image of an object in front of a convex miror is reduced. Therefore, the correct answer is (A).

Answer:

A. Reduced

5) Let's first find the image distance from the mirror equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}},$$ $$d_i=\dfrac{1}{-\dfrac{1}{50\ cm}-\dfrac{1}{30\ cm}}=-18.75\ cm.$$

The sign minus means that the image is virtual and located behind the mirror.

Let's apply the magnification equation and find the magnification of the mirror and height of the image:

$$M=-\dfrac{d_i}{d_o}=-\dfrac{-18.75\ cm}{30\ cm}=0.63,$$ $$M=\dfrac{h_i}{h_o},$$ $$h_i=Mh_o=0.63\times10\ cm=6.3\ cm.$$

The image is upright and 6.3 cm long. Therefore, the correct answer is (B).

Answer:

B. 6.3 cm long and upright

6) An object closer to a converging lens than its focal point always has an image that is virtual. Therefore, the correct answer is (B).

Answer:

B. Virtual

7) We can find the image distance from the lens equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{d_o}},$$ $$d_i=\dfrac{1}{\dfrac{1}{12\ cm}-\dfrac{1}{10\ cm}}=-60\ cm.$$

The sign minus means that the image is virtual. Therefore, the correct answer is (D).

Answer:

D. -60 cm