A wire is bent in the shape of a right-angled | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The wire is placed between $f/2$ and $f$. For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Let a point $P$ on

Vedclass · Accepted Answer

Vedclass · Answer

(B) The wire is placed between $f/2$ and $f$. For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Let a point $P$ on the hypotenuse be at a distance $x$ from the focus $f$,so $u = -(f-x)$.Substituting this into the mirror formula: $\frac{1}{v} - \frac{1}{f-x} = -\frac{1}{f}$.$\frac{1}{v} = \frac{1}{f-x} - \frac{1}{f} = \frac{x}{f(f-x)}$.Thus,$v = \frac{f(f-x)}{x}$.The magnification $M = -\frac{v}{u} = -\frac{f(f-x)/x}{-(f-x)} = \frac{f}{x}$.For the vertical leg of the triangle at $u = -f/2$,the image is at $v = -f$ (real,inverted,magnified by $2$).For the point at the focus $(u = -f)$,the image is at infinity $(v \rightarrow \infty)$.As $x$ increases from $0$ to $f/2$,the magnification $M = f/x$ decreases from $\infty$ to $2$. The image of the hypotenuse is a curve that extends to infinity,consistent with option $B$.

Column $I$	Column $II$
$A$. Concave Mirror	$(p)$ Real image
$B$. Convex Mirror	$(q)$ Virtual image
$C$. Convex Lens	$(r)$ Magnified image
$D$. Concave Lens	$(s)$ Image at infinity

$A$ wire is bent in the shape of a right-angled triangle and is placed in front of a concave mirror of focal length $f$,as shown in the figure. Which of the figures shown in the four options qualitatively represent$(s)$ the shape of the image of the bent wire? (These figures are not to scale.)

Similar Questions

When seen in green light,the saffron and green portions of our National Flag will appear to be

Correct exposure for a photographic print is $10 \, s$ at a distance of $1 \, m$ from a point source of $20 \, cd$. For an equal fogging of the print placed at a distance of $2 \, m$ from a $16 \, cd$ source,the necessary time for exposure is.......$s$.

If the polarising angle of a piece of glass for green light is $54.74^o$,then the angle of minimum deviation for an equilateral prism made of the same glass is......$^o$ (Given $\tan 54.74^o = 1.414$)

$A$ satisfactory photographic print is obtained when the exposure time is $10 \, s$ at a distance of $2 \, m$ from a $60 \, cd$ lamp. The time of exposure required for the same quality print at a distance of $4 \, m$ from a $120 \, cd$ lamp is $... \, s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module