4. An object is 30 cm from a thin, positive lens. The focal length of the lens is 10 cm. Where will the image be formed by this positive lens? q = Is the image real of imaginary? Mark the position of this image on the figure. 5. There is a negative lens to the right of the positive lens, at a distance of L = 10 cm away. What is the distance between this negative lens and the image formed by the positive lens? d = 6. The image formed by the positive lens is the object of the negative lens. Is the image formed by the positive lens to the right or the left of the negative lens? In this case, is the object distance positive or negative? To the negative lens, p = 7. The focal length of the negative lens is also 10 cm (negative). Calculate where the image will be formed by the negative lens. q = mark the position of this image on the figure. Is the image real or virtual? For a diverging lens, the focal length is negative by definition. Thus in the thin lens equation, if the object distance is positive, then the image distance will always be negative. How did we make the diverging lens to form a real image?
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5 \text{ cm} \] So, the image formed by the positive lens is 7.5 cm to the right of the lens. Show more…
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The focal length of a diverging lens is negative. If f = -23 cm for a particular diverging lens, where will the image be formed of an object located 32 cm to the left of the lens on the optical axis? ______ cm to the left of the lens? What is the magnification of the image? b. A small object is placed to the left of a convex lens and on its optical axis. The object is 29 cm from the lens, which has a focal length of 12 cm. Determine the location of the image formed by the lens. (Enter your answer in cm from the lens.) ______ cm from the lens Describe the image. (Select all that apply.) real virtual upright inverted enlarged reduced on the left side of the lens on the right side of the lens
Madhur L.
The telephoto lens shown consists of a converging lens of focal length $+6.0 \mathrm{~cm}$ placed $4.0 \mathrm{~cm}$ in front of a diverging lens of focal length $-2.5 \mathrm{~cm} .(a)$ Locate the image of a very distant object. (b) Compare the size of the image formed by this lens combination with the size of the image that could be produced by the positive lens alone. ( $a$ ) If the negative lens were not employed, the intermediate image $A B$ would be formed at the focal point of the $+6.0$ lens, $6.0 \mathrm{~cm}$ distant from the $+6.0$ lens. The negative lens decreases the convergence of the rays refracted by the positive lens and causes them to focus at $A^{\prime} B^{\prime}$ instead of $A B$. The image $A B$ (that would have been formed by the $+6.0$ lens alone) is $6.0-4.0=2.0 \mathrm{~cm}$ beyond the $-2.5$ lens and is taken as the (virtual) object for the $-2.5$ lens. Then $s_{o}=-2.0 \mathrm{~cm}$ (negative because $A B$ is virtual), and $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{-2.5 \mathrm{~cm}}-\frac{1}{-2.0 \mathrm{~cm}}=\frac{1}{10 \mathrm{~cm}} \quad \text { or } \quad s_{i}=+10 \mathrm{~cm} $$ The final image $A^{\prime} B^{\prime}$ is real and $10 \mathrm{~cm}$ beyond the negative lens. (b) Magnification by negative lens $=\frac{\overline{A^{\prime} B^{\prime}}}{\overline{A B}}=-\frac{s_{i}}{s_{o}}=-\frac{10 \mathrm{~cm}}{-2.0 \mathrm{~cm}}=5.0$ so the diverging lens increases the magnification by a factor of $5.0 .$ Notice that the magnification produced by the convex lens is negative and so the net magnification of both lenses is negative: the final image is inverted.
As shown an object is placed $40 \mathrm{~cm}$ in front of a converging lens that has $f=+8.0 \mathrm{~cm} .$ A plane mirror is $30 \mathrm{~cm}$ beyond the lens. Find the positions of all images formed by this system. For the lens $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8.0}-\frac{1}{40}=\frac{4}{40} \quad \text { or } \quad s_{i}=10 \mathrm{~cm} $$ This is image $A^{\prime} B^{\prime}$ in the figure. It is real and inverted. $A^{\prime} B^{\prime}$ acts as an object for the plane mirror, $20 \mathrm{~cm}$ away. A virtual image $C D$ is formed $20 \mathrm{~cm}$ behind the mirror. Light reflected by the mirror appears to come from the image at $C D$. With $C D$ as object, the lens forms an image of it to the left of the lens. The distance $s_{i}$ from the lens to this latter image is given by $$ \frac{1}{s_{i}}=\frac{1}{f}-\frac{1}{s_{o}}=\frac{1}{8}-\frac{1}{50}=0.105 \quad \text { or } \quad s_{i}=9.5 \mathrm{~cm} $$ The real images are therefore located $10 \mathrm{~cm}$ to the right of the lens and $9.5 \mathrm{~cm}$ to the left of the lens. (This latter image is upright.) A virtual inverted image is found $20 \mathrm{~cm}$ behind the mirror.
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